I am trying to solve $$x^n-1=a^m$$ equation in integers for odd n.
For the case $a=2$:
From the relation: $$(x-1)(x^{n-1}+x^{n-2}+...+1)=2^m$$
we can see that $x$ should odd, and the $n$ is even. So for odd $n$ it has no solution.
Now assume that the $a>2$ the question is if there is a solution for given odd $n$. Wondering if this is known topic and there is any paper about this.