I get $x=\pm 3$ and $x = \pm 45$ and that's it. It is likely that the set of solutions is finite, see PILLAI'S CONJECTURE There is a fine description of how Pillai's conjecture would be written by @ShreevatsR in comment below:
that "for fixed positive integers A,B,C the equation $Ax^n−By^m=C$ has
only finitely many solutions $(x,y,m,n)$ with $(m,n) \neq (2,2).$"
Here with $(A,B,C)=(1,1,23)$ it says that $x^n−y^m=23$ has finitely
many integer solutions $(x,y,m,n).$ The OP's claim is that it has
infinitely many integer solutions of the form $(2,x,2,n).$ This is
Meanwhile, I can describe how to rapidly exhaust possible solutions, by my methods. We know that $n$ must be odd in $x^2 - 2^n = -23.$ So, take $n= 2t+1$ and make a new variable, $y= 2^t.$ The result is $$ x^2 - 2 y^2 = -23. $$ The seed values are $(x,y) = (3,4)$ and $(x,y) = (-3,4).$ We want all solutions such that $y$ turns out to be a power of 2. Now, given a solution $(x,y),$ we get all possible solutions by repeatedly taking the result of applying an element of the automorphism group/isometry group/orthogonal group of $x^2 - 2 y^2,$ namely $$ (3x-4y,-2x+3y). $$
Now, $y=4$ is a power of 2, so that is a start, with $x=\pm 3$
The first string is $$ (3,4), (-7,6), (-45,32),(-263,186),(-1533,1084),(-8935,6318), \ldots $$
So this one gives $$ (x = \pm 3, y = 4), \; \; \; (x = \pm 45, y = 32) $$ as successes
The other string is
$$ (-3,4),(-25,18),(-147,104), (-857,606),(-4995,3532),(-29113,20586), \ldots $$
So you can see how I became skeptical about there being any more solutions with $y$ a power of 2.
Note that Erick Wong has pointed out a proof as a simple application of elliptic curves.
Meanwhile, I am now awake, see KATY PERRY.