Diophantine equation $n^2+23=2^{2k}$

Question

Find all solutions (if any) to $n^2+23=2^{2k}$. I've tried computing mod low integers but found nothing contradictory.

Answer 1

$$n^2+23=2^{2k}\implies(2^k+n)(2^k-n)=23\implies2^k+n=23\text{ and }2^k-n=1\implies2^{k+1}=24$$

(Ah, I didn't notice quasi's comment below the OP until after posting. Or Ivan's!)

Answer 2

If $n$ is not divisible by $3$, the number $n^2+23$ is divisble by $3$, hence it cannot be a power of $2$

If $n$ is divisible by $3$, we have $n^2+23\equiv 2\mod 3$, hence $n^2+23$ cannot be a perfect square.

Hence, the equation has no integral solutions.

Answer 3

Since $2^{2k}-\left(2^k-1\right)^2=2^{k+1}-1$, for $k\ge4$, the difference between consecutive squares is at least $31$. Thus, $2^{2k}-23$ cannot be a square for $k\ge4$.

For $k\le2$, $2^{2k}-23\lt0$, so it cannot be a square.

This leaves $k=3$ and $2^6-23=41$, which is not a square.

Thus, there are no solutions to $2^{2k}-23=n^2$.