if $k > 1$ and $m\ge 1$ are such that $a^k = (2^m-1)(2^m+1)$, then does $a | 2^m-1$ or $a| 2^m+1$
Let $a$, $k$ and $m$ be positive integers with $k\geq2$ and $$a^k=(2^m-1)(2^m+1).$$ The two factors on the right hand side are coprime, because they are both odd and their difference is $2$. Then by unique factorization, both are $k$-th powers. That is, there exist positive integers $b$ and $c$ such that $$2^m-1=b^k\qquad\text{ and }\qquad 2^m+1=c^k.$$ Subtracting the former from the latter shows that $$c^k-b^k=2,$$ but of course there are no two positive powers that differ by $2$.
So the answer is vacuously yes, because there are no such integers.