The theorems of Sylow are very well known and almost every mathematician learns in his undergraduate course.
The applications of Sylow theorems given in books are of the kind
"If $|G|=....$ then show that $G$ is not simple/ $G$ is solvable/ ..."
I would like to know if there are some other, interesting, applications of this theorem.
I know exactly one other direct application of the Sylow theorems outside of group theory, which is to proving the fundamental theorem of algebra.
Suppose $K$ is a Galois extension of $\mathbb{R}$. We'll aim to show that either $K = \mathbb{R}$ or $K = \mathbb{C}$. (In particular, $\mathbb{C}$ itself must therefore be algebraically closed.) Let $G$ be its Galois group and let $H$ be the Sylow $2$-subgroup of $G$.
By Galois theory, $K^H$ is an odd extension of $\mathbb{R}$. But $\mathbb{R}$ has no nontrivial odd extensions: any such extension has primitive element something with an odd degree minimal polynomial over $\mathbb{R}$, but any such polynomial has a root by the intermediate value theorem. Hence $K^H = \mathbb{R}$, or equivalently $H = G$, so $G$ has order a power of $2$.
But now $K$ is an iterated quadratic extension of $\mathbb{R}$, and it's easy to explicitly show using the quadratic formula that the only nontrivial quadratic extension of $\mathbb{R}$ is $\mathbb{C}$, which itself has no nontrivial quadratic extensions.
The fundamental theorem of algebra is probably the best example, but how about the following proof that a finite subgroup of the multiplicative group of a field is cyclic:
Let $G \subset F^\times$ be finite. Let $H_p$ be a $p$-Sylow subgroup. Then if $|H_p|=p^k$, we claim that $H_p$ is simply the set of all roots in $F$ of the polynomial $x^{p^k}-1$: every element of $H_p$ is a root by LaGrange's theorem, and there can be no more roots since the degree of the polynomial is $p^k$. Now using this, it's easy to see that $H_p$ is cyclic: it's generated by any $y \in H_p$ such that $y^{p^{k-1}}\neq 1$ (such $y$ exist because there are only $p^{k-1}$ roots of $x^{p^{k-1}}-1$). But now since $G$ is abelian, $G$ is the direct product of its Sylow subgroups (this is where a Sylow theorem is used - although you could also use abelian group theory...), which are all cyclic of relatively prime order.
Let $p$ be a natural number larger than $1$. Then $p$ is prime if and only if $(p-1)! \equiv -1 \mod p$.
This is Wilson's theorem. It can be proved using Sylow's theorem.
Indeed the implication $\Leftarrow$ is rather elementary, for the implication $\Rightarrow$, if $p$ is a prime the symmetric group $G=S_p$ of degree $p$ contains exactly $(p-1)!$ elements of order $p$ ($p$-cycles): this is also elementary. Each of them generates a Sylow $p$-subgroup of $G$, because the order of the Sylow $p$-subgroups of $G$ is exactly $p$.
It follows that the number of Sylow $p$-subgroups of $G$ is $(p-1)!/(p-1)= (p-2)!$, since each Sylow $p$-subgroup of $G$ contains precisely $p-1$ elements of order $p$ and any two Sylow $p$-subgroups of $G$ intersect trivially.
Sylow's theorem then implies that $(p-2)! \equiv 1 \mod p$, and the result follows by multiplying both sides by $p-1$.
The Sylow Theorems often play a crucial role in finding all groups of a certain order. For example, all groups of order $pq$, or all groups of order $p^n$, where $p$ and $q$ are primes can be found in this manner. You may find more information in this book by J.S. Milne, chapter 5.