The solution consists of three steps:

1. $2^{m-1}-1$ is divisible by 127.
2. $2^{m-1}-1$ is divisible by $m$.
3. 127 and $m$ are coprime.

For 1: It holds that $2^{6} \equiv 1 \bmod 7$. From $p \equiv 1 \bmod 6$ it follows that $2^{p} \equiv 2 \bmod 7$, thus $7 \mid m-1$. With $2^{7} \equiv 1 \bmod 127$ it follows that $2^{m-1} \equiv 1 \bmod 127$.
For 2: It holds that $2^{p} \equiv 2 \bmod p$ (Fermat's little theorem), i.e., $p \mid m-1$. From $2^{p} \equiv 1 \bmod m$ it follows that $2^{m-1} \equiv 1 \bmod m$.
For 3: It holds that $2^{p} \equiv 1 \bmod 127$ if and only if $p$ is divisible by 7 (write $p=7 k+r$ with $0 \leq r<7$). If $r>0$, this is not the case, so $m$ is not divisible by 127, and since 127 is a prime number, 127 and $m$ are coprime.
Remarks: The relationship $\operatorname{gcd}\left(2^{a}-1,2^{b}-1\right)=2^{\operatorname{gcd}(a, b)}-1$ allows for a shortening of the solution. Instead of congruence arithmetic, the divisibility statement $a-b \mid a^{n}-b^{n}$ can be used in some cases. Some participants used an incorrect version of the Euler-Fermat theorem: it states $a^{\phi(m)} \equiv 1 \bmod m$ for $\operatorname{gcd}(a, m)=1$, but only for $m$ a prime number does $\phi(m)=m-1$ hold.