2^N-1=kN; where k is an integer
2^N=kN+1 => 2^N is relatively prime to N => 2 is relatively prime to N.
By Fermat's Little Theorem 2^N=2 modN (for N>1).
From this we can say that 2^N-1 can't be multiple of N for N>1
So, only possible solution is N=1.