Turkish Mathematical Olympiad

Let us show that both $m$ and $127$ divide $2^{m-1}-1$. By Fermat's little theorem $2^p\equiv 2(\mathrm{mod}p)$, so $m=2^p-1\equiv 1(\mathrm{mod}p)$, which implies $p\mid m-1$. Therefore, $2^p-1\mid 2^{m-1}-1$, so $m\mid 2^{m-1}-1$.

On the other hand, $6\mid p-1$ so $63=2^6-1\mid 2^{p-1}-1$. This implies $7\mid 2^p-2$, so $7\mid m-1$. Therefore, $127=2^7-1\mid 2^{m-1}-1$.

We complete the solution by showing that $m$ and $127$ are relatively prime. Since $127$ is prime, it is enough to show that $m$ is not divisible by $127$. Let

$p=7k+n$ ($0\le n<7$). $k>1\Rightarrow p>7$ and $p$ is not divisible by $7\Rightarrow n\ne 0$. Now $127=2^7-1\mid 2^{7k}-1\Rightarrow 127\mid 2^{7k+n}-2^n=2^p-2^n$. If $127\mid m$, then $127\mid 2^n-1$ which is impossible since $0<2^n-1<127$. Done.