The equation can be rewritten as $(x-N)(y-N)=N^2$. So all positive solutions come from the factorizations of $N^2$ as a product $uv$, and all such factorizations come from a solution. Just put $x=N+u$, $y=N+v$.
Thus the problem of finding the solutions of your equation is essentially equivalent to the problem of finding all factors of $N^2$, which is easily solved if we know all the factors of $N$.
However, for very large $N$, finding all the factors of $M$ can be computationally challenging.
Remarks: Let $N$ have prime power factorization $N=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$. Then the number of ordered pairs $(u,v)$ of positive integers such that $uv=N^2$ is equal to $(2a_1+1)(2a_2+1)\cdots (2a_k+1)$. This is also the number of solutions of your equation. Thus you can manufacture examples where the number of solutions is large.