Is there an easy way to determine when this fractional expression is an integer? For $x,y\in \mathbb{Z}^+,$ when is the following expression an integer? 
$$z=\frac{(1-x)-(1+x)y}{(1+x)+(1-x)y}$$
The associated Diophantine equation is symmetric in $x, y, z$, but I couldn't do anything more with that. I tried several factoring tricks without luck. The best I could do was find three solutions such that $0<x\le y\le z$. They are: $(2,5,8)$, $(2,4,13)$ and $(3,3,7)$.
The expression seems to converge pretty quickly to some non-integer between 1 and 2.

We write out a solution, to check whether something like it can be "borrowed" for a problem set.
The numerator is $1-x-y-xy$, which can be written as $-[(x+1)(y+1)-2]$. The denominator is $-[(x-1)(y-1)-2]$. So our fraction is 
$$\frac{(x+1)(y+1)-2}{(x-1)(y-1)-2}.$$
We can take $x=1$ or $y=1$. The ratio is then an integer, albeit negative. This gives infinitely many trivial solutions. We will list the non-trivial solutions.
By symmetry we can assume that $x \le y$. Let $x=2$. Then we want $(-1-3y)/(3-y)$ to be an integer. But
$$\frac{3y+1}{y-3}=3+\frac{10}{y-3}.$$
So $y-3$ must divide $10$, giving (since $y \ge x$), the solutions $y=2$, $4$, $5$, $8$, and $13$.
Next we deal with $x=3$. A calculation similar to the previous one (but shorter) gives that the only $y\ge 3$ are given by $y=3$ and $y=7$.
Next we deal with $x=4$. We need $\dfrac{5y+3}{3y-5}$ to be an integer. Any common divisor of these two numbers must divide $3(5y+3)-5(3y-5)$, which is $34$. The only divisor of $34$ which is of the form $3y-5$, where $y \ge 4$, is given by $y=13$.  
Similarly, if $x=5$, we get the solution $y=8$.
For the rest, we use essentially the analysis of @Ragib Zaman. One needs to verify that $2x+2y< xy-x-y-1$, or equivalently that $(x-3)(y-3)>10$. This is true if $x\ge 6$ and $y\ge 7$. (It fails at $x=y=6$, but that is not a solution.)  
