The problem comes from a problem I encountered when I wrote the article

Find all positive integer $m$ such $$2^{m}+1\mid5^m-1$$ it seem there no solution. I think it might be necessary to use quadratic reciprocity knowledge to solve this problem. If $m$ is odd then $2^m+1$ is divisible by 3 but $5^m-1$ is not. so $m$ be even, take $m=2n$, then $$4^n+1\mid25^n-1.$$ if $n$ is odd,then $4^n+1$ is divisible by $5$, but $25^n-1$ is not so $n$ is even,take $n=2p$, we have $$16^p+1\mid625^p-1.$$