Peter wrote: Determine all triples of positive integers $ (a, m, n)$ such that $ a^m + 1$ divides $ (a + 1)^n$. http://www.kalva.demon.co.uk/short/soln/sh00n4.html

If $ a,m > 1$ and $ (a,m) \neq (2,3)$, by Zsigmondy's theorem there is a prime $ p$ dividing $ a^m + 1$ but not $ a + 1$. Thus there are no solutions. Trivially checking the other cases, we get that all other cases always are solutions, regardless what $ n$ is, the only exception being $ (a,m) = (2,3)$ where we need $ n \geq 2$.