Hong Kong competition problems

By Dirichlet's theorem, there are infinitely many primes $p$ of the form $13k + 1$. We claim that we can set $m = p$ whenever $p > 8191$.

Firstly, $p$ divides $2^{p-1} - 1$ by the Fermat little theorem. Secondly, we have
$$2^{p-1} - 1 = 2^{13k} - 1 = (2^{13} - 1)(2^{13(k-1)} + 2^{13(k-2)} + \dots + 1),$$
which is divisible by $2^{13} - 1 = 8191$. As $p$ and $8191$ are relatively prime, the number $\frac{2^{p-1}-1}{8191p}$ is an integer.