Is there an integer $m\geq 1$ such that $2^m+3^m$ is a perfect power?



${2}^{p}+{3}^{p}={a}^{n}$ , then n=1 for any p ?
If you really wanted to prove this (and I'm afraid that I'm not sure why you would), you could invoke a Theorem of Darmon and Merel for $n=2$ and $3$, check that there are no solutions with $p \leq 5$, say, and then write down the usual $(n,n,n)$ Frey curve, assuming $n \geq 5$ is prime (which leads to a weight $2$, level $6$ cuspidal newform and hence the desired contradiction).


Of course, this is an absurdly big hammer for such a problem and likely something {\it much} simpler works.


