First, we will prove a lemma.
Lemma Let $p$ be a prime number satisfying $p \equiv 3 (mod \ 4)$. Then,
$$
p\mid a^2+b^2 \implies p\mid a \ \text{and} \ p\mid b.
$$
Proof of Lemma First, note that if $p\mid a$, then $p$ must divide $b$ and vice versa. So, assume $p\nmid a,b$. Now,
$$
\frac{a^2}{b^2}\equiv -1 (mod p)
$$
gives us a contradiction, since $-1$ is not a quadratic residue modulo $p$, if $p-3$ is divisible by $4$ (well-known fact from number theory).
Equipped with this, we will attack the problem. Suppose,
$$
\frac{x^2+y^2}{x-y} = p\alpha
$$
for some $\alpha$ and $p \equiv 3 \ (mod \ 4)$. Then, we can define $x = px'$ and $y = py'$ to argue that
$$
\frac{x'^2+y'^2}{x'-y'} = \alpha.
$$
With this reasoning, starting from any $p\alpha$, we can arrive at either $\alpha = 1$ or $5$, which is the only prime divisor of $1995$, that is congruent to $1$ modulo $4$.
Case 1 $\frac{x'^2 + y'^2}{x'-y'}=1$. With some very basic inequalities, you can convince yourself that this is not possible.
Case 2 $\frac{x'^2 + y'^2}{x'-y'}=5$. In this case, note that
$$
x'^2 \leq x'^2 + y'^2 = 5x'-5y' \leq 5x' \implies x'\leq 5.
$$
With some algebra, the only solutions are $(x',y') = (3,1)$ and $(2,1)$. Now from here, you can get the entire solution set. For instance, to make $\frac{x^2+y^2}{x-y}=5$, the pairs above work. To make it, say, $3\times 5 \times 7$, multiply these pairs by $3\times 7$. This produces the entire set, and note that you have $16$ possibilities, as $1995$ has 16 divisors.
