Mathematical Reflections J339: Solving $\frac{x-1}{y+1} + \frac{y-1}{z+1} + \frac{z-1}{x+1} = 1$ in positive integers I have been trying to solve $\frac{x-1}{y+1} + \frac{y-1}{z+1} + \frac{z-1}{x+1} = 1$ in positive integers and I'm shaky on one specific step. I solved the problem through case-by-case analysis and I wonder if someone else has a better, more natural solution.
Observe that (2,2,2) is a trivial solution.
Assume that one variable, WLOG $x=1$. Then we need to solve $\frac{y-1}{z+1} + \frac{z-1}{2} = 1$ which after clearing denominators is equivalent to $2(y-1) + z^2-1 = 2z+2$ or $(z-3)(z+1)=-2(y-1)$. The quadratic is even whenever $z$ is odd so testing $z=3$ and $z=1$ shows that $(1,1,3)$ (and its permutations) are solutions.
This is my shaky step: Besides $(1,1,3)$, all other solutions have $x,y,z\geq2$.
I then verify that other than $(2,2,2)$ there are no solutions with all three of $x,y,z\geq2$.
Is this logic correct? Can I say that no other solutions involve 1 now that I have done casework on WLOG $x=1$?
Thanks in advance!

observe that the quadratic $(z-3)(z+1)$ has a minimum where $z=1$ and its value there is $-4$, so we get $-2(y-1)\ge -4$, that is $y\le 3$. since $y\ge 1$, we get $(z-3)(z+1)\le 0$, so $z\le 3$ too.
