Prove that if $(2^m-1) \mid (2^n-1)$ then $m \mid n$.
Since $m \ge 1$, we can use the Division Algorithm (1.7) to write
$$n=mq+r$$
where $q$ and $0 \le r < m$ are unique.
Let's write $2^n-1$ as follows
$$ \begin{align} 2^n -1 & = (2^{qm+r} - 2^r) + (2^r - 1) \\ \\ & = 2^r(2^{qm} - 1) + (2^r -1) \end{align}$$
By definition $m \mid qm$ and so by Proposition (4.9) we have $(2^m-1) \mid (2^{qm}-1)$. This means, for some integer $k$, we have $k(2^m-1) = (2^{mq}-1)$. And so
$$ 2^n -1 = k2^r(2^{m} - 1) + (2^r -1) $$
If we assume $(2^m-1) \mid (2^n-1)$, then
$$ (2^m-1) \mid (2^n-1) \implies (2^m-1) \mid k2^r(2^{m} - 1) + (2^r -1) $$
This means $2^m-1 \mid 2^r-1$.
But from the division algorithm $r < m$, and so $2^r-1 < 2^m-1$. The only way $2^m-1 \mid 2^r-1$ is if $2^r-1=0$. This means $r=0$. And so $n=mq+0 = mq$. This gives u the desired conclusion
$$ (2^m-1) \mid (2^n-1) \implies m \mid q $$