prove that there infinitely many $m$ such that $\frac{2^{m-1}-1}{8191m} \in \mathbb{N}$

Question

Here is what i did : $$2^{13}\equiv 1 [8191]$$ $$2^{26}\equiv 1 [8191]$$ since $m$ is odd so $$m-1=26k$$ with $$k\in \mathbb{N}$$ and we have $$2^{m-1}-1\equiv 0 [m]$$
Hence $m$ is prime soooo $$m=26k+1$$ with $m$ prime I think that i should prove that there infinitely many primes $m$ such that$m\equiv 1[26]$

Answer 1

What saulspatz said in the comments is true, $2^{m-1}\equiv 1\pmod m$ does not imply that $m$ is prime. However, it is true that $2^{p-1}\equiv 1\pmod p$, for any prime $p>2$, by Fermat's little theorem.

Thus, you are correct, since you only need to show that there are infinitely many such $m$, it is enough to show that there are infinitely many primes of the form $26k + 1$. This follows from Dirichlet's theorem on arithmetic progressions.