Since $ \frac{a^k-1}{a-1}=\sum_{j=0}^{k-1}a^j\tag{1} $ we immediately get that $q^m-1\;|\;q^n-1$ when $m|n$.
Now, suppose that $q^m-1\;|\;q^n-1$ and that $n=km+r$ where $0\le r< m$. Then $ \begin{align} \frac{q^n-1}{q^m-1} &=\frac{q^{km+r}-q^{km}}{q^m-1}+\frac{q^{km}-1}{q^m-1}\\ &=q^{km}\frac{q^r-1}{q^m-1}+\frac{q^{km}-1}{q^m-1}\\ &\in\mathbb{Z}\tag{2} \end{align} $ It immediately follows from $(1)$ that $\frac{q^{km}-1}{q^m-1}\in\mathbb{Z}$. Therefore, we must also have $q^{km}\frac{q^r-1}{q^m-1}\in\mathbb{Z}$: $ q^m-1\;|\;q^{km}(q^r-1)\tag{3} $
Since $\left(q^{km},q^m-1\right)=1$, $(3)$ implies that $q^m-1\;|\;q^r-1$. However, since $0\le r< m$, we have that $0\le q^r-1< q^m-1$. Therefore, $q^r-1=0$; that is, $r=0$ and $n=km$; hence, $m|n$.