This should be $(2^m-1)^2\mid 2^n-1$. Let $n=mk$ where $2^m-1\mid k$. Then, $2^n-1=(u-1)(u^{k-1}+\cdots+1)$ for $u=2^m\equiv 1\pmod{2^m-1}$, and therefore $u^{k-1}+\cdots+1\equiv k\equiv 0\pmod{2^m-1}$. All in all, $(2^m-1)^2\mid 2^n-1$. Now let $(2^m-1)^2\mid 2^n-1$. Then using $2^m-1\mid 2^n-1$ we obtain $n=mk$ for some $k$ (a well-known fact). Then setting $u=2^m$, we find $2^n-1 = u^k-1=(u-1)(u^{k-1}+\cdots+1)$, so $2^m-1\mid u^{k-1}+\cdots+1$. As $u^{k-1}+\cdots+1\equiv k\pmod{2^m-1}$, we must further have $2^m-1\mid k$, so $m(2^m-1)\mid n$ as desired.