Clearly $a\leq b$. Assume $a<b$. Let $p$ be a prime number greater than $b$ and $n=(a+1)(p-1)+1$. By, FLT, $a^n\equiv a(\mod p)$ and $b^n\equiv b(\mod p$). Now $n\equiv-a(\mod p)$. Then, $a^n+n\equiv a+n\equiv0(\mod p)$. So, $b^n+n\equiv 0(\mod p)$. However, $b^n+n\equiv b-a(\mod p)$ and $0<b-a<b<p$. Hence, a contradiction. So, $a=b$.

This is ISL 2005 N6. It can be generalised to proving that $a^n+f(n)$ divides $b^n+f(n)$ for all sufficiently large $n$ if and only if $a=b$, where $f$ is any non-constant integer polynomial.