<s>[INST] An Interesting Number Theory Problem: Solve $2520=(x+y+xy)^2+2xy+2y-3x, x, y \in \mathbb Z$ Solve:

$2520=(x+y+xy)^2+2xy+2y-3x$

Please help, I tried Solving SFFT and Quadratic, but couldn't end up with a result.
Edit: I solved this much.I factorized it
$((x+1)(y+1)-1)^2+2((x+1)(y+1)-1)-5x$
Then substituted $t=((x+1)(y+1)-1)$
so $t^2+2t-5x=2520$
and after this i am stuck
Edit 2: Solve for Integers
 [/INST] Denote $z=x+1, w=y+1$ then your result $$t^2+2t-5x=2520 \implies (t+1)^2-5(z-1)=2519 \\\implies w^2z^2-5z=2516=z(w^2 z - 5) \implies z|2516=2^2\cdot 17\cdot 37$$
If $z$ is even then $w^2z-5$ is odd, so $4|z, z=4u, 4u^2w^2=5u+629 \implies u\equiv -1 \pmod 4$, and $u|17\cdot 37$, therefore $u=-1, -17, -37, -629$, none of which yields a perfect square for $5u+629$.
If $z$ is odd then $wz$ is odd, $(wz)^2 \equiv 1 \pmod 4$, so $z\equiv 1 \pmod 4, z=1, 17, 37, 629$. We have the following four cases:
$z=1, w^2-5=2516$, no solution;
$z=17, 17w^2-5=148 \implies w=\pm 3 \implies x=16, y=2 \text{ or } -4;$
$z=37, 37w^2-5=68$, no solution;
$z=629, 629w^2-5=4$, no solution.
So the only solutions are $(16, 2)$ and $(16, -4)$.
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