Find all $m$ such $2^m+1\mid5^m-1$

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 $$16^p+1\mid625^p-1.$$

Here is a proof.

Theorem. $$2^m+1$$ never divides $$5^m-1$$.

Assume that there is some $$m$$ such that $$2^m+1$$ divides $$5^m-1$$. We already know that $$m$$ must be divisible by $$4$$. Let $$m = 2^n a$$ with an odd integer $$a$$ and $$n \ge 2$$. The $$n$$th Fermat number $$F_n = 2^{2^n} + 1$$ is congruent to $$2$$ mod $$5$$ (this uses $$n \ge 2$$), so it has a prime divisor $$p$$ such that $$p \equiv \pm 2 \pmod 5.$$ We know that $$p-1 = 2^{n+1}k$$ for some integer $$k$$. Since $$\left(\frac{5}{p}\right) = \left(\frac{p}{5}\right) = -1$$, we have that $$5^{2^n k} = 5^{(p-1)/2} \equiv -1 \pmod p,$$ so $$5^{mk} = (5^{2^n k})^a \equiv -1 \pmod p$$ as well. In particular, $$5^m \not\equiv 1 \pmod p.$$ On the other hand, $$2^m = (2^{2^n})^a \equiv (-1)^a = -1 \pmod p.$$ Thus $$p$$ divides $$2^m+1$$, but does not divide $$5^m-1$$, a contradiction.

• @Joe Thanks for improving the layout! Dec 14, 2018 at 16:58

There might well be a very elementary argument for this, but in the spirit of taking a hammer to a fly, one can prove that the number of $$m$$ such that $$2^m+1 \mid 5^m-1$$ is finite by invoking a theorem of Bugeaud, Corvaja and Zannier [Math. Z. 2003] which implies, in this context that, given $$\epsilon > 0$$, $$\gcd (4^m-1, 5^m-1) \leq e^{\epsilon m}$$ for suitably large $$m$$. Schmidt's Subspace Theorem is used here, so the result is ineffective.