If $x^2 - y^2 = 1995$, prove that there are no integers x and y divisible by 3 
If $x^2 - y^2 = 1995$, prove that there are no integers $x$ and $y$ divisible by $3$.

I'm quite new to this. As in the title I would like to ask if my reasoning here is correct and if my answer would be considered acceptable.


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*If $x$ or $y$ is divisible by $3$, but not both, then:
$\begin{align}x \equiv \pm 1 \pmod{3} \land y \equiv 0 \pmod{3} 
&\iff x^2 \equiv 1 \pmod{3} \land y^2 \equiv 0 \pmod{3}
\\
&\iff x^2 - y^2 \equiv 1 \pmod{3}\end{align}$
So if $x$ or $y$ is divisible by $3$ but not both then no values for $x$ and $y$ will ever make $x^2 - y^2$ divisible by $3$.


*If $x$ and $y$ are divisible by $3$:


Since both $x$ and $y$ are divisible by $3$ they can be expressed $x = 3m$ and $y = 3n$ for some integers $m$, $n$.
$(3m)^2 - (3n)^2 = 1995 \iff 9m^2 - 9n^2 = 1995 \iff 3(m^2-n^2) = 665$
And it's easy to see that there are no whole number solutions for m and n, and thus there are no whole number solutions for x and y such that $x^2 - y^2 = 1995$ where x or y (or both) is divisible by 3.
Is this a reasonable proof? Are there more elegant ways to do it?

Your argument is fine and well expressed. More succinctly you could paraphrase parts of your argument as follows.
If precisely one of $x,y$ is divisible by $3$ then $x^2-y^2$ is not divisible by $3$.
If both of $x,y$ are divisible by $3$ then $x^2-y^2$ is divisible by $9$.
