This cropped up in an otherwise simple-looking problem. Find the solutions for $a, b, n \in \mathbb{Z}$ and $b, n > 1$ for the Diophantine equation:
$b^n + 1 = a^2$
Alternatively:
$a^2 - b^n = 1$
One can see that if $n$ is even there are no solutions. But for $n$ odd, there can be solutions, one of which is of course evident in $3^2 - 2^3 = 1$
Is this an open problem or do we know the solutions here?
Edit: Is there a simple, elementary solution to this special case?