Not so quick resolution

Write $n=pq$ for district primes $p,q$ . Then the equality becomes $p^3q^3+x^2=y^2,$ or $p^3q^3=(y+x)(y-x).$ If $p$ and $q$ are both odd then any choice of factor of $p^3q^3$ for $y+x$ will yield integer solutions for $x,y$ . There are 16 (positive) factors, but since $y+x$ must be larger than $y-x$ (since we are interested in positive solutions), there are eight positive integer solutions when $n$ is odd.

When $p=2$ , in order to find integer solutions we must have both $y+x$ , $y-x$ even. There are 8 such (positive) factors, yielding four positive integer solutions when $n$ is even.

Thus, the answer is $4+8=\boxed{12}$ .

Let $n = p_{1} p_{2}$ where $p_{2}, p_{2}$ prime.

Thus we have the desired equality can be written as $(y-x)(y+x) = p_{1}^{3} p_{2}^{3}$ . It is obvious that $x$ is a positive integer if and only if $(y-x) < (y+x)$ and $x, y \in N$ if and only if $(y-x) and (y+x)$ have the same parity.

(i) $p_{1} = 2$

Therefore $(y-x)(y+x) = 2^{3} p_{2}^{3}$ . $(y-x) \equiv (y+x) (mod 2)$ if and only if $2 | (y-x)$ and $2 | (y+x)$ .

Therefore, there are exactly 8 ordered pairs of $((y-x), (y+x)$ .

By bijection there are exactly 4 ordered pairs which satisfies $(y-x) < (y+x)$ .

Therefore there are excatly 4 ordered positive integers $(x, y)$ that satisfies the above equality.

(ii) $p_{1} \geq 3$

Therefore, $(y-x)$ and $(y+x)$ have the same parity if and only if $2 \nmid (y-x)$ ans $2 \nmid (y+x)$ , which all pairs of $((y-x), (y+x))$ satisfies.

It is obvious that $n^{3}$ have exactly 16 positive divisor.

Therefore, by bijection there are exactly 8 pairs of $((y-x), (y+x))$ with (y-x) < (y+x) that satisfies the original equality.

Therefore there are exactly 8 pairs of positive integers (x, y) such that the above equality is satisfied.

Therefore the are exactly 12 pairs of positive integers (x, y) such that the above equality is satisfied.