Prove $q^m -1 \mid q^n-1$ implies $m\mid n$
This is the standard proof of the fact. Basically suppose the opposite: $q^m-1$ divides $q^n-1$ but $m$ does not divide $n$. And assume that $n$ is the smallest possible for this $m$. Then first, $n>m$ otherwise $q^n-1\le q^m-1$.
Second, divide $m$ by $n$ with a remainder as in your question, etc. and produce smaller $n$ for the same $m$, a contradiction. Subtracting $q^m-1$ is just a trick to achieve this goal. There is no deep meaning here.