First of all, the proof is correct and I congratulate you on the excellent effort. I will only offer a few small comments on the writing.
It's not clear until all the way down at (5) that you intend to do a proof by contradiction, and even then you never make it explicit. It's generally polite to state at the very beginning of a proof if you plan to make use of contradiction, contrapositive, or induction.
Tiny detail, maybe even a typo: $n$ and $m$ are integers, not necessarily naturals, so the statement at the end of (2) needs to reflect that. But for integers, also $x>-1$ implies $x\geq 0$, so it's not a big deal.
You didn't really need to make $n$ and $m$ positive since the only place you use positivity is at the very, very end you need $m>0$ to derive the contradiction. You don't even use it when you multiply by $d$ since that relied on the expressions being positive and not the individual numbers themselves. This is the only place I can imagine really see simplifying the proof.
As it stands, it would make the reader more comfortable if you named the positive versions of $(n,m)$ as $(N,M)$ or $(n',m')$ or something.
Finally, as you hint at, you don't need to consider the positive-positive case. But perhaps you should be more explicit why this is, earlier in the proof.
Best Answer
For any real numbers $a$ and $b\neq 0$, we have $-\frac{a}{b}=\frac{-a}{b}=\frac{a}{-b}$. Also, $-\frac{-a}{b}=-\frac{a}{-b}=\frac{-a}{-b}=\frac{a}{b}$.