I have written a proof in the following format as part of my abstract reasoning unit, just wondering if there's any faults in my logic.
Let $m \equiv 2 \operatorname{mod} 4$, then $\nexists x,y \in \mathbb N : x^2 - y^2 = m$.
Proof (by contradiction):
Assume $\exists x,y \in \mathbb N : x^2 - y^2 = m$, namely,
$$x^2 - y^2 \equiv 2 \operatorname{mod} 4$$
Thus, using the theorem reduction modulo m:
$$x = mk + r : k,m,r,x \in \mathbb N$$
Therefore,
$$x^2 - y^2 = 4k + 2 : k \in \mathbb N$$
$$x^2 - y^2 = 2(2k+1)$$
Hence, $2 \operatorname | x^2 - y^2$.
However, since $x^2 - y^2 \equiv 2 \operatorname{mod} 4$, we know this can't be true since $2$ is a common factor of $4$. Meaning when dividing by $2$ there will also be some remainder, say $j\in \mathbb N : x^2 - y^2 \equiv j \operatorname{mod} 2$. This is a contradiction, hence it was wrong to assume the original statement was wrong. Thus the original statement is true.
$\square$
You writing is a bit hard to understand to be fair. Anyway how dividing with $2$ leads you to a contradiction. You need to divide $x^2-y^2$ by $2$ also, which you haven't done. Also I don't see a way of doing this.
A better solution would be to notice that $x^2-y^2 = (x+y)(x-y)$ and note that $2\mid (x-y)(x+y)$. If $2 \mid x-y$. However then $2 \mid x-y+2y = x+y$. This would mean that $4 \mid (x+y)(x-y) = x^2 -y^2$, which is a contradiction. Now similarly you can discard the case when $2 \mid x+y$ and obtain the wanted contradiction.