nt.number theory – find all $ m $ such $ 2 ^ m + 1 | 5 ^ m-1 $

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

Find all the whole numbers $ m $ such
$$ 2 ^ {m} +1 | 5 ^ m-1 $$
It seems that there is no solution, I think it might be necessary to use the knowledge of quadratic reciprocity to solve this problem.
Yes $ m $ it's strange then $ 2 ^ m + 1 $ It is divisible by 3 but $ 5 ^ m-1 $ It is not.
so $ m $ be even take $ m = 2n $,so
$$ 4 ^ n + 1 | 25 ^ m-1 $$
Yes $ n $ it's strange, then $ 4 ^ n + 1 $ It is divisible by $ 5 $,but $ 25 ^ m-1 $ It is not
so $ n $ it's even, take $ n = 2p $,we have
$$ 16 ^ p + 1 | 625 ^ m-1 $$