<s>[INST] Find all $x,y$ so that $\dfrac{x+y+2}{xy-1}$ is an integer. I am trying to find the integers $x,y$ so that 
$\dfrac{x+y+2}{xy-1}$ is an integer. 
What I have done:
I suppose there exists $t$ such that $$t=\dfrac{x+y+2}{xy-1}$$ where $xy\neq 1$ then consider the following scenarios:
$$x=y$$ $$x>y>0$$  $$x>0>y$$ ,etc.
This approach  helps me find the solutions but it is very long. Any simpler method?
 [/INST] If $x=y=-1$, then $x+y+2=xy-1=0$. Therefore, some of the solutions below need to be prefaced with "except for $x=y=-1$".

If $x=y$, then $\frac{x+y+2}{xy-1}=\frac{2}{x-1}$, which says that
$$
x=y\in\{0,2,3\}\quad\text{are the only solutions with $x=y$}\tag{1}
$$
Furthermore,
$$
x+y=-2\quad\text{is a solution}\tag{2}
$$
We also have that
$$
xy=0\quad\text{is a solution}\tag{3}
$$
If $x=-1$ or $y=-1$, $\frac{x+y+2}{xy-1}=-1$. Therefore,
$$
(x+1)(y+1)=0\quad\text{is a solution}\tag{4}
$$
If $y=1$, then the solutions to $\frac{x+y+2}{xy-1}=\frac{x+3}{x-1}\in\mathbb{Z}$ must have $-3\le x\le5$. Checking gives
$$
\{(-3,1),(-1,1),(0,1),(2,1),(3,1),(5,1)\}\tag{5}
$$
If $x=1$, then the solutions to $\frac{x+y+2}{xy-1}=\frac{y+3}{y-1}\in\mathbb{Z}$ must have $-3\le y\le5$. Checking gives$$
\{(1,-3),(1,-1),(1,0),(1,2),(1,3),(1,5)\}\tag{6}
$$
Since
$$
|xy-1|-|x+y+2|\ge(|x|-1)(|y|-1)-4\tag{7}
$$
if $x+y\ne-2$ and $(|x|-1)(|y|-1)\ge5$, then $(x,y)$ cannot be a solution. Since all the solutions with $|x|\le1$ or $|y|\le1$ are covered above, we only need to check $2\le|x|,|y|\le5$. Checking these $64$ cases, the only new solutions not counted above are
$$
\{(2,5),(5,2)\}\tag{8}
$$
Therefore, all solutions are given by $(1)$-$(6)$ and $(8)$.
</s>