\(k=2^p-1\\ N=k^2-1\\ N=(2^p-1)^2-1\\ N=2^{2p}-2*2^p+1-1\\ N=2^p*2^p-2*2^p\\ N=2^p(2^p-2)\\ N=2^{p+1}(2^{p-1}-1)\\ so\\ 2^{p+1}\text{ is a factor of }N\)
So I have shown that 2^(p+1) is a factor of N
Which is the other way around from what you asked.
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