Finding all positive integer solutions for $x+y=xyz-1$ How do I manually solve $x+y=xyz-1$ assuming that $x, y$ and $z$ are positive integers? I was able to guess all possible solutions, but I do not know how to show that these are the only ones:
$x=1, y=1, z=3$
$x=1, y=2, z=2$
$x=2, y=1, z=2$
$x=2, y=3, z=1$
$x=3, y=2, z=1$  
Any hints would be appreciated.

We have
$$z=\frac{x+y+1}{xy}=\frac 1y+\frac 1x+\frac{1}{xy}\le 1+1+1$$$$\Rightarrow z=1,2,3$$
Also,
$$zxy-x-y=1\iff z^2xy-zx-zy+1=z+1\iff (zx-1)(zy-1)=z+1$$
So, for $z=1$, we have $(x-1)(y-1)=2$.
For $z=2$, we have $(2x-1)(2y-1)=3$.
For $z=3$, we have $(3x-1)(3y-1)=4$.
