If $2^k - 1$ is prime then $k$ is prime, using $2^{m} -1 \vert 2^{mn} - 1$

Question

Prove if $2^k - 1$ is prime, then $k$ is prime.

using the fact that for positive integers $m,n$:

$2^m -1 \vert 2^{mn} - 1$

so $2^{mn} - 1 = (2^m - 1)k$ some $k$ (**)

for contrapositive, suppose $k = k_1k_2$

now we w.t.s. $2^{k_1k_2} - 1$ is composite as well

Can I show this using (**) letting $m,n$ be $k_1,k_2$? And that's all I need to do ?