Consider a general ring and count the number of solutions to $x^2=1$. The answer is either $2^n$ for some $n\in \mathbb{N}$ or $\infty$.
Is this correct? For some examples it's correct, $\mathbb{Z}^n$ has $2^n$ solutions. $\mathbb{Z}/n\mathbb{Z}$ is discussed here. Can we find counterexamples or prove the statement?
This is an extension of the comment by rso.
The number of involutions in $M_n(\mathbb{F}_2)$ for $n \geq 0$ is the sequence starting with $1,1,4,22,316$, which is the OEIS sequence A053722 (and clearly these don't have to be powers of $2$). You can verify the value $22$ for $n=3$ via brute force with this GAP code.
# Number of solutions of x^2 = 1 in a given ring
NumberOfSolutions := function(ring)
local elements, x, solutions;
elements := Elements(ring);
solutions := Filtered(elements, x -> x^2 = One(ring));
return Length(solutions);
end;
p := 2;
n := 3;
MatrixRing := FullMatrixAlgebra(GF(p), n);
Print(NumberOfSolutions(MatrixRing)); # 22
Since we are in a ring of characteristic $2$, the equation $X^2=1$ is equivalent to $(X-1)^2=0$. Hence, the number is also the number of solutions of the equation $X^2 = 0$, which is much easier to handle.
But actually it is much easier to work with $M_n(\mathbb{F}_p)$ where $p$ is an odd prime. Then $X^2=1$ is equivalent to $\bigl(\frac{X+1}{2}\bigr)^2=\frac{X+1}{2}$ as mentioned in a comment by Daniel, so that it is sufficient to count the number of solutions to $X^2=X$, which are the idempotent elements. For their enumeration (also for similar questions), see sections 1.10 and 2.4 in
Kent E. Morrison, Integer Sequences and Matrices Over Finite Fields (pdf)
For $p=3$ we get the sequence A053846 which starts as $1, 2, 14, 236$.
To have this result, you need to ensure that this set is a group (you need stability by multiplication) if so this is direct since it is a $\mathbb{F}_2$-vector space. But this is equivalent to the commutativity of this set since if it is commutative it is a group, and if it is a group, $xyx^{-1}y^{-1} = xyxy = (xy)^2 = 1$.
Now you have a very easy (not so much in fact) counterexample by considering $$R = \mathbb{F}_3 \cdot \langle a,b \mid a^2=1 , b^2= 1, ab = -ba\rangle.$$ It is a free $\mathbb{F}_3$-algebra of three variables quotiented by the bisided ideal generated by $$(a^2-1,b^2-1,ab + ba).$$
Now the only square roots of $1$ in $R$ are $$\pm 1,\pm a,\pm b,\pm(a+b+ab), \pm (a-b+ab), \pm (a+b-ab), \pm(a-b-ab)$$ which means you have $14$ elements which is not a power of 2.
To check them you just need to see that this has a dimension of $4$ and a base is given by $(1,a,b,ab)$, so every element can be represented by a tuple $(l,m,n,p)$ and we have $$(l,m,n,p)^2 = (l^2+m^2+n^2-p^2,2lm,2ln,2lp)$$ Now for this to be 1 we need
Case 1: $l\neq 0$ then $m=n=p=0$ which means $l = \pm 1$.
Case 2: $l = 0$ then $m^2 + n^2 = p^2+1$ this gives you all the non trivial solutions (either we have only one on the left that is non zero, or we have all of them nonzero)
NB: $(ab)^2 = -1$, $aab = b$, $aba = -b$, $bab=-a$, $abb = a$.
Consider the ring $M_2(\mathbb{F}_q)$ for $q$ a power of a prime other than 2. Then by standard linear algebra, the elements with $x^2 = 1$ correspond bijectively to the pairs $(S, T)$ of subspaces of $\mathbb{F}_q^2$ with $\mathbb{F}_q^2 = S \oplus T$, i.e. $S$ and $T$ are complementary subspaces. (In one direction, take the eigenspaces of $\lambda = 1$ and $\lambda = -1$; in the other direction, take the unique automorphism which sends $s + t \mapsto s - t$ for $s\in S, t\in T$. Note that it is important here that we're working over a field of characteristic not equal to 2, which is necessary to conclude that $t^2-1$ is square-free so the minimal polynomial of $x$ is also square-free, implying that $x$ is diagonalizable.)
Now we just use some combinatorics to count the number of such pairs. If $\dim S = 0$, $S$ is the zero space and $T$ is the full space, which gives one choice; similarly, if $\dim S = 2$, then $S$ is the full space and $T$ is the zero space, which gives one choice. If $\dim S = 1$, then there are $\frac{q^2-1}{q-1} = q+1$ subspaces of dimension 1 to choose for $S$, and then $q$ subspaces of dimension 1 to choose for $T$ (since you just have to choose a subspace of dimension 1 which is not the same as $S$). In conclusion, there are $2+(q+1)q = q^2+q+2$ elements $x \in M_2(\mathbb{F}_p)$ with $x^2 = 1$. On the other hand, $q^2+q+2$ is not necessarily a power of 2; for example, $3^2+3+2 = 14$, so $M_2(\mathbb{F}_3)$ has 14 square roots of 1.
I had basically the same approach as the one in jules' answer. So this won't add much new, but I will post it anyway, and maybe it motivates the counterexample a bit more.
Let $G := \{x \in R: x^2 = 1\}$. If $R$ is commutative, this is an abelian group under multiplication such that $x^2=1$ for every $x \in G$. Such groups are called elementary $2$-abelian and are the same as vector spaces over $\mathbb{F}_2$. If $d$ is the dimension, we get an isomorphism $G \cong \mathbb{F}_2^d$ and the claim follows.
More generally, if $R$ is not necessarily commutative, but $G$ is multiplicatively closed, i.e. $x^2=y^2=1 \implies (xy)^2=1$, then $G$ is a subgroup, and it is abelian, so the same proof works.
So to look for a counterexample, we need to have (at least) a ring with elements $x,y$ with $x^2=1$, $y^2=1$, but $(xy)^2 \neq 1$. The universal example of a ring with two elements whose squares are $1$ is the ring $$R := \mathbb{Z}\langle X,Y \rangle / (X^2-1, Y^2-1),$$ which is isomorphic to the group ring $\mathbb{Z}[G]$ where $G = C_2 * C_2 \cong D_{\infty}$ (infinite dihedral group). The elements of $G$ are $1$, $X$, $Y$, $XY$, $YX$, $XYX$, $YXY$, etc. and these form the $\mathbb{Z}$-basis of $R$. But I had a really hard time to calculate the solutions of $r^2=1$ here, even though it is quite likely that $\pm 1, \pm X, \pm Y$ are the only (six) solutions, unless I am missing something. (Can someone check this?)
The ring becomes considerably easier to handle if we introduce another relation, such as $YX=-XY$, so that $X,Y$ anti-commute. The quotient ring $$R := \mathbb{Z} \langle X,Y \rangle / (X^2-1,Y^2-1,XY+XY)$$ has only $4$ generators as a $\mathbb{Z}$-module: $1$, $X$, $Y$, $XY$. In this ring, we can compute $$(a+bX+cY+dXY)^2 = (a^2 + b^2 + c^2 - d^2) + 2ab X + 2ac Y + (2ad) XY.$$ This is $1$ iff we have $ab=ac=ad=0$ and $a^2+b^2+c^2-d^2=1$. If $a=0$, this means $b^2+c^2-d^2=1$. If $a \neq 0$, we get $b=c=d=0$ and $a^2=1$, i.e. $a = \pm 1$.
Now, unfortunately, $b^2+c^2-d^2=1$ has infinitely many solutions in $\mathbb{Z}$. But we can easily replace $\mathbb{Z}$ by any other commutative ring $k$. As soon as $k$ is an integral domain with $2 \in k^{\times}$, the same arguments work with the $k$-algebra $$R := k\langle X,Y \rangle / (X^2-1,Y^2-1,XY+YX).$$ The equation $r^2=1$ has the two trivial solutions $\pm 1$ as well as the solutions $bX + cY + dXY$, where $b^2+c^2-d^2=1$. The latter equation is equivalent to $b^2 = 1 + d^2 - c^2$. Counting the solutions is now doable for every finite field in particular, for the case $k = \mathbb{F}_3$ see jules' answer. We just need to take $c,d \in k$ and check if $1 + d^2 - c^2$ is a square. Then there are $2$ solutions for $b$ (resp. $1$ if the square is zero).