In a previous question Can a square number plus 1 become a square? the questioner asked in passing at the end whether $a^n+1$ could ever be a perfect square (for $a \ge 1$ and $n \ge 2$). I provided a partial answer demonstrating that $a^n+1 \neq m^2$ when $a$ is odd. In looking at $a$ being even, I got bogged down.
Here is how far I got. $a^n+1 = m^2$ implies $a^n=(m+1)(m-1)$. If $a=2^kb$ is even (with $b$ being odd), then both $(m+1)$ and $(m-1)$ are even. Two consecutive even numbers have only one factor of $2$ in common, i.e. $\gcd {\frac {(m+1)}{2},\frac {(m-1)}{2} =1}$.
$a^n=2^{kn}b^n$ and one of the factors ($(m+1)$ and $(m-1)$) can be represented as $2c^n$ and the other factor can be represented as $2^{kn-1}d^n$, $c$ and $d$ both odd. The difference between the factors is just $2$, but we don't know a priori which factor is the larger, so we can say that $2c^n-2^{kn-1}d^n= \pm 2$. Dividing through by $2$ and rearranging we get $c^n \pm 1=2^{kn-2}d^n$.
I mention here that for $c=d=k=1, n=3$ we get $1+1=2$ deriving from the well known example $2^3+1=3^2$.
I can't find a discussion of this in previously posted questions, and I can't see a way forward from this point.
My question is: Other than the one example cited, does the relationship $c^n \pm 1=2^{kn-2}d^n$ have solutions in the integers?
Added by edit: Since my question arises from a previous question whether $a^n+1$ can be a perfect square, we can see that $n$ must be odd, since $a^{2j}+1=(a^j)^2+1$ which cannot be a perfect square since no two integer squares differ by $1$. Hence in the context of my question, where the exponent $n$ is the same as in the original question, it must be the case that $n=2j+1$.