Does the equation $(xy+1)(xy+x+2)=n^2$ have a positive integer solution? Does there exist a positive integral solution $(x, y, n)$ to $(xy+1)(xy+x+2)=n^2$? If there doesn't, how does one prove that?

It looks that Vieta jumping helps.
For fixed positive integer $y$ choose a minimal positive integer $x$ for which $(xy+1)(xy+x+2)$ is a perfect square.
Denote $4(xy+1)(xy+x+2)=4n^2=(2xy+x+3-z)^2$ for some integer $z=2n-2xy-x-3$, this yields $0<z<x+3$ and rewrites as $z^2-2z(2xy+x+3)+x^2+2x+1=0$.  Note that $x$ must divide $z^2-6z+1$, for each $z\leqslant 5$ this gives several variants for $x$ for which it is straightforward to check that $y$ does not appear to be a positive integer. If $z\geqslant 6$, we may replace $x$ to $x'=(z^2-6z+1)/x>0$ (which is another root of the same quadratic equation in $x$.) This contradicts to the minimality since $z^2-6z+1<(z-3)^2<x^2$. 
Remark: for the new pair $(x',y)$ we have different value of $z$, as $2x'y+x'+3-z$ becomes negative, but it is still true that $(x'y+1)(x'y+x'+2)$ is a perfect square.
