1

So far I'm just starting on this, and I'm suspecting that there is no N>1. 

Observations:
If there is such an N, then it must be true that:
2^N - 1 = k*N   where k is an integer

Also, 2^N - 1 in binary is a series of N 1's, which suggests that
2^N - 1  is divisible by 2^m - 1  if N is divisible by m
Not sure if this helps.

Now I'll pause and wait for someone else to present an elegant proof.