The Cayley-Hamilton theorem is equivalent to: Let $R$ be a ring and let $M_n(R)$ be $n\times n$ matrices over $R$. Then the minimal polynomial of $A \in M_n(R)$ over $R$ divides the characteristic polynomial of $A$.
For instance. In order to reduce the confusion of having $X = $ a matrix in a polynomial. Let $R'$ be the subring of matrices $\{ a I : a \in R\}$. It's clearly isomorphic to $R$. Now consider the characteristic polynomial as an element of $R'[X]$.
Here is my proof of the Cayley Hamilton theorem. I'll share the intuition behind it first:
Intuition in a Nutshell: For any endomorphism $\Phi : V \rightarrow V$, we have a factorization of the determinant $\text{det}(\Phi) I$ into the adjugate and the matrix itself: $$ \text{det}(\Phi) I = \text{adj}(\Phi) \circ \Phi$$ We want to use this to get a factorization of the characteristic polynomial $p(t)$ of $\phi$ into some polynomial analogous to the adjugate and a linear term $t - \phi $: $$p(t) = f(t)(t - \phi)$$ These two factorizations are analogous, and in fact, if we get the formality right, we can view these as corresponding factorizations in isomorphic rings $\text{End}(V \otimes k[t])$ and $\text{End}(V) \otimes k[t]$.
Let's try to work this out a little more formally. The main question is, what are the two isomorphic rings I mentioned in which these are corresponding factorizations?
Let $V$ be a finite dimensional vector space over a field $k$. One of the rings is $\text{End}_k(V)[t] = \text{End}_k(V) \otimes k[t]$. The characteristic polynomial $p(t)$ of $\phi \in \text{End}_k (V)$ naturally lives in $\text{End}_k(V)[t]$ from the natural map $\text{End}_k(V) \rightarrow \text{End}_k(V) \otimes_k k[t]$. In other language, we view $t \text{Id}_V - \phi $ as having endomorphisms as coefficients, and then take the determinant, which is then in $\text{End}_k (V)[t]$.
The other ring is $\text{End}_{k[t]}(V \otimes_k k[t])$. $\Phi := 1 \otimes t - \phi \otimes 1 $ is an element in this ring, and we have a factorization $\text{det}(\Phi) 1_{V \otimes_k k[t]} = \text{adj}(\Phi) \Phi$.
In the isomorphism $$\text{End}_k ( V \otimes_k k[t]) \cong \text{End}_k (V)[t]$$ We have corresponding elements $$\Phi \leftrightarrow t - \phi$$ and $$\text{det}(\Phi) \leftrightarrow p(t)$$ Therefore, the factorization $\text{det}(\Phi) 1_{V \otimes_k k[t]} = \text{adj}(\Phi) \Phi$ corresponds to a factorization $p(t) = f(t)(t-\phi)$ in $\text{End}_k (V) [t]$. And that's the whole idea!
If you want a more formal version, and a construction of the claimed isomorphism, read on!
Theorem: Let $V$ be a finitely generated $k$-module. If $\phi : V \rightarrow V$ is a $k$-linear map, then the evaluation homomorphism $\text{ev}_{\phi} : k[t] \rightarrow \text{End}_k (V)$ sends the characteristic polynomial $\text{char}(\phi)$ to $0$.
Let's start by constructing an isomorphism $F : \text{End}_{k} (V)[t] \rightarrow \text{End}_{k[t]} (V \otimes_k k[t])$ as follows. We have isomorphisms $$\text{End}_{k[t]} (V \otimes_k k[t]) \cong \text{Hom}_k (V, \text{Hom}_{k[t]}(k[t], V \otimes_k k[t])) \cong \text{Hom}_k(V, V \otimes_k k[t])$$
These isomorphisms can be established by creating canonical maps in both directions and showing that they are inverse to each other. Now we have a canonical map in a single direction,
$$\text{End}_k (V) \otimes_k k[t] \rightarrow \text{Hom}_k(V, V \otimes_k k[t])$$
sending $\phi \otimes t^n$ to the map sending $v$ to $\phi(v)t^n$. This is injective, and surjective since $V$ is finitely generated. Composing these isomorphisms gives an isomorphism $F : \text{End}_{k} (V)[t] \rightarrow \text{End}_{k[t]} (V \otimes_k k[t])$.
Now we argue as before. View $t - \phi$ as a $k[t]$-linear endomorphism of $V \otimes_k k[t]$. Under the isomorphism $F$, $\text{char}(\phi)$ maps to $\text{det} (t - \phi) 1_{V \otimes_k k[t]} )$ and $F ( t - \phi ) = t - \phi$. $t - \phi$ divides $\text{det}(t - \phi) 1_{V \otimes_k k[t]}$ in $\text{End}_{k[t]} (V \otimes_k k[t])$, since $\text{det} (t - \phi) 1_{V \otimes_k k[t]} = \text{adj}(t - \phi) (t - \phi)$, where $\text{adj}(t - \phi)$ is the adjugate matrix. Therefore, $t - \phi$ divides $\text{char}(\phi)$ in $\text{End}_{k}(V)[t]$. So $\text{char}(\phi)$ has $\phi$ as a root in $\text{End}_k(V)$, so that the evaluation homomorphism $\text{ev}_{\phi} : k[t] \rightarrow \text{End}_k (V)$ sends the characteristic polynomial $\text{char}(\phi)$ to $0$.
This has no pretense to be "The" answer!
I am no algebraist but I remember a nice proof I was taught when I was I student, I thought I'd share it.
In a nutshell: True for diagonalizable matrices, then use "algebraic continuation".
Let's write down some details.
Lemma ("algebraic continuation"): Let $k$ be an infinite field. Let $P, Q \in k[X_1, \dots, X_n]$ be polynomials of $n$ variables with $Q \neq 0$ . If $P$ vanishes on the set $\{x \in k^n ~\colon~ Q(x) \neq 0\}$, then $P = 0$.
This lemma expresses that "non-empty open sets are dense in the Zariski topology". It's not hard to show (I'll give you a hint if you want).
Now:
Theorem (Cayley-Hamilton): Let $R$ be a commutative unital ring. Let $A \in M_n(R)$ be a square matrix and denote by $\chi_A(X) \in R[X]$ its characteristic polynomial. Then $\chi_A(A)$ is the zero matrix.
Let's give a proof when $R = k$ is an infinite field for the moment. By the "algebraic continuation" lemma, it is enough to show that the theorem is true when $A$ lies in some "dense open set". More precisely, each coefficient of the matrix $\chi_A(A)$ is a polynomial in the $n^2$ coefficients of $A$. It is enough to show that it vanishes on some set $\{Q \neq 0\}$, where $Q$ is a nonzero polynomial in $n^2$ variables. Let's take $Q(A) = \mathrm{Disc}(\chi_A)$ (the discriminant of the polynomial $\chi_A$). The set where $Q \neq 0$ consists precisely of matrices $A$ whose eigenvalues are all distinct in an algebraic closure $\bar{k}$ of $k$. Such matrices are diagonalizable over $\bar{k}$ so it is easy to check that $\chi_A(A)$ = 0 (I'll let you do that).
We're done!
Wait, how does this extend to an arbitrary unital ring? Well, each of the coefficients of the matrix $\chi_A(A)$ is actually a polynomial in the $n^2$ coefficients of $M$ with integer coefficients. These polynomials must be zero because we showed that Cayley-Hamilton holds for $\mathbb{Q}$ hence $\mathbb{Z}$. (NB: I think some people would say something like "$\mathbb{Z}$ is an initial object in the category of unital rings" or whatever).
Let $R$ be a commutative unital ring and $M$ be a finite-rank free unital $R$-module. Let $$ a\colon M\to M $$ be an endomorphism of $M$ and $\chi_a\in R[X]$ be the characteristic polynomial of $a$.
Consider the left action $(\triangleleft_a)$ of $\operatorname{End}_R(M)[X]\cong\operatorname{End}_{R[X]}(M[X])$ on $\operatorname{End}_R(M)$ defined by the rules:
Then: $$ (X - a)\triangleleft_a\operatorname{id}_M =\operatorname{id}_Ma - a\operatorname{id}_M = a - a = 0, $$ and \begin{align*} \chi_a(a) &=\chi_a\triangleleft_a\operatorname{id}_M\\ &=\operatorname{det}(X - a)\triangleleft_a\operatorname{id}_M\\ &=\operatorname{adj}(X - a)(X - a)\triangleleft_a\operatorname{id}_M\\ &=\operatorname{adj}(X - a)\triangleleft_a((X - a)\triangleleft_a\operatorname{id}_M)\\ &=\operatorname{adj}(X - a)\triangleleft_a0\\ &= 0. \end{align*}
Details here: https://arxiv.org/abs/2105.09285.
Here is another presentation of the same idea.
First, observe that, by abuse of notation (or by some "identifications"), \begin{align*} \operatorname{id}_M\otimes X - a\otimes 1 &\in\operatorname{End}_R(M)[X] =\operatorname{End}_R(M)\otimes_R R[X] \cong\operatorname{End}_{R[X]}(M[X]),\\ \operatorname{adj}(\operatorname{id}_M\otimes X - a\otimes 1) &\in\operatorname{End}_R(M)[X] =\operatorname{End}_R(M)\otimes_R R[X] \cong\operatorname{End}_{R[X]}(M[X]), \end{align*} and $$ \chi_a =\operatorname{det}(\operatorname{id}_M\otimes X - a\otimes 1) =\operatorname{adj}(\operatorname{id}_M\otimes X - a\otimes 1) (\operatorname{id}_M\otimes X - a\otimes 1). $$
Consider the four $R$-module homomorphisms defined as follows: \begin{align*} \operatorname{End}_R(M)\otimes_R R[X] \otimes_R\operatorname{End}_R(M)\otimes_R R[X] &\to\operatorname{End}_R(M)\otimes_R R[X]\\ f_1\otimes P_1\otimes f_2\otimes P_2&\mapsto f_1f_2\otimes P_2P_1,\\[2ex] \operatorname{End}_R(M)\otimes_R R[X] &\to\operatorname{End}_R(M)\\ f\otimes P&\mapsto fP(a),\\[2ex] \operatorname{End}_R(M)\otimes_R R[X] \otimes_R\operatorname{End}_R(M)\otimes_R R[X] &\to\operatorname{End}_R(M)\otimes_R R[X]\otimes_R\operatorname{End}_R(M)\\ f_1\otimes P_1\otimes f_2\otimes P_2&\mapsto f_1\otimes P_1\otimes f_2P_2(a),\\[2ex] \operatorname{End}_R(M)\otimes_R R[X]\otimes_R\operatorname{End}_R(M) &\to\operatorname{End}_R(M)\\ f\otimes P\otimes g&\mapsto fgP(a). \end{align*}
It is easy to check that these $R$-module homomorphisms form the commutative diagram $$\require{AMScd} \begin{CD} \operatorname{End}_R(M)\otimes_R R[X] \otimes_R\operatorname{End}_R(M)\otimes_R R[X] @>>>\operatorname{End}_R(M)\otimes_R R[X]\\ @VVV @VVV\\ \operatorname{End}_R(M)\otimes_R R[X]\otimes_R\operatorname{End}_R(M) @>>>\operatorname{End}_R(M) \end{CD} $$
Now, consider the image in $\operatorname{End}_R(M)$ of $\operatorname{adj}(\operatorname{id}_M\otimes X - a\otimes 1)\otimes (\operatorname{id}_M\otimes X - a\otimes 1)$ under the mappings of this diagram.
When going through $\operatorname{End}_R(M)\otimes_R R[X]$, we have: $$ \operatorname{adj}(\operatorname{id}_M\otimes X - a\otimes 1)\otimes (\operatorname{id}_M\otimes X - a\otimes 1) \mapsto\chi_a\mapsto\chi_a(a). $$ When going through $\operatorname{End}_R(M)\otimes_R R[X]\otimes_R\operatorname{End}_R(M)$, we have: $$ \operatorname{adj}(\operatorname{id}_M\otimes X - a\otimes 1)\otimes (\operatorname{id}_M\otimes X - a\otimes 1) \mapsto\operatorname{adj}(\operatorname{id}_M\otimes X - a\otimes 1)\otimes 0 = 0 \mapsto 0. $$
