Is my proof correct? $2^n=x^2+23$ has an infinite number of (integer) solutions. This is how I tried to prove it. Is it correct? Thanks!!
$2^n = x^2+23$
$x^2$ must be odd, therefore $x^2 = 4k+1$, where $k \in \mathbb{N}$.
$2^n=4k+24$
$k=2(2^{n-3}-3)$
Since $x^2=4k+1$,$ \ \ \ \ $ $k_1 = \frac{x_1^2-1}{4}$
and $k_2=\frac{(x_1+2)^2-1}{4}=\frac{x_1^2+4x_1+3}{4}$
If we substitute $x_1^2=4k_1 + 1$, we end up with:
$k_2=k_1 + \sqrt{4k_1+1} + 1$
Therefore, finding solutions of $k=2(2^{n-3} - 3)$ is comparable to
finding the solutions of $k+\sqrt{4k+1}+1=2(2^{n-3}-3)$
Let $p=2(2^{n-3}-3)$
Therefore, $(\sqrt{4k+1})^2=(p-k-1)^2$, so
$(p-k)^2 = 2(p+k)$
Since $p$ has an infinite number of solutions $(p-k)^2=2(p+k)$ also has an infinite number of solutions, which implies  the the original does also.

Here is another heuristic argument along probabilistic lines that there are finitely many solutions.
The distance between perfect squares near $n$ is approximately $2\sqrt{n}$. Thus, the probability of a given integer $n$ to be a perfect square is approximately $\frac1{2\sqrt{n}}$. Summing the probability that $2^n-23$ is a perfect square gives
$$
\begin{align}
\sum_{n=5}^\infty\frac1{2\sqrt{2^n-23}}
&\le\frac16+\sum_{n=6}^\infty\frac1{2\sqrt{2^{n-1}}}\\
&=\frac16+\frac{\sqrt2+1}8
\end{align}
$$
According to the Borel-Cantelli Lemma, this indicates that the probability that there are finitely mny solutions is $1$.
Of course, this really proves nothing, because the same argument suggests that there are finitely many perfect squares of the form $2^n$, when in fact $2^n$ is a perfect square whenever $n$ is even. However, in the absence of proof to the contrary, this provides a decent guess.
