Solution:

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.

On 1: We have $2^{6} \equiv 1 \bmod 7$. From $p \equiv 1 \bmod 6$ it follows that $2^{p} \equiv 2 \bmod 7$, hence $7 \mid m-1$. With $2^{7} \equiv 1 \bmod 127$ it follows from this that $2^{m-1} \equiv 1 \bmod 127$.

On 2: We have $2^{p} \equiv 2 \bmod p$ (Fermat's little theorem), i.e. $p \mid m-1$. From $2^{p} \equiv 1 \bmod m$ it then follows that $2^{m-1} \equiv 1 \bmod m$.

On 3: We have $2^{p} \equiv 1 \bmod 127$ if and only if $p$ is divisible by 7 (write $p=7k+r$ with $0 \leq r<7$ and use $2^{7} \equiv 1 \bmod 127$). Since $p>7$, this is not the case, so $m$ is not divisible by 127, and since 127 is a prime number, 127 and $m$ are coprime.