Find $x ,y$ satisfying : $x,y$ are $2$ positive integers and $(xy+x+2)(xy+1)$ is a perfect square. Find $x ,y$ satisfying : $x,y$ are $2$ positive integers and $(xy+x+2)(xy+1)$ is a perfect square.
I have solved this problem and will post the solution as soon as possible. Hope everyone can check my solution! Thanks very much !

For a start:
Let $z=xy+1$, then $$z(z+x+1)=a^2\implies \boxed{z^2+z(x+1)-a^2=0}$$
Since discriminant is perfect square we have $$d^2 =(x+1)^2+4a^2$$ which means that (Pythagorean triples) $$d=k\cdot (u^2+v^2),\;\; x+1 =k\cdot(u^2-v^2),\;\; a = k\cdot uv$$ for some positive integers $u,v,k$ and $\gcd(u,v)=1$. Since $x+1>0$ we have $u>v$. Then (solving boxed equation) $$xy+1 =k\cdot {v^2-u^2\pm (u^2+v^2)\over 2}$$
Since $z$ is positive we have only $+$ choise, so we have $xy+1 = kv^2$ so $y = {kv^2-1\over k(u^2-v^2)-1}$


 ??? Since for every integer $m$ we have $ \gcd(mk+1,k)=1$ thus $k(u^2-v^2)-1\mid k(2v^2+u^2) \implies k(u^2-v^2)-1\mid 2v^2+u^2$ and so
 $$k(u^2-v^2)-1\leq  2v^2+u^2$$

So we have only finite possibilities for $k$ ...

