Find all triplets $(x,y,z)\in\mathbb{R}^3$ such that \begin{align*} x^2-yz &= |y-z|+1, \\ y^2-zx &= |z-x|+1, \\ z^2-xy &= |x-y|+1. \end{align*}
Solution: Interchanging $x,y,z$ by $(-x,-y,-z)$ we may w.l.o.g. assume that $x+y+z \ge 0$. This is symmetric in $x,y,z$ so we may w.l.o.g. suppose that $x \ge y \ge z$ so that the system reads as \begin{align*} x^2-yz&=y-z+1\\ y^2-zx&=x-z+1\\ z^2-xy &=x-y+1 \end{align*}Subtracting the second from the first equation we find that $(y-z)(x+y+z-1)=0$ so that $y=z$ or $x+y+z=1$. Subtracting the third from the second equation we find that $(x-y)(x+y+z+1)=0$ so that $x=y$ or $x+y+z=-1$. Since clearly $x=y=z$ is not a solution, we must have $x=y$ and $x+y+z=1$. Hence $x=-2y-1$ and the first equation is $x^2-y^2=1$ so that we find $(2y+1)^2-y^2=1$ and hence $y=0$ or $y=-\frac{4}{3}$. However $y=0$ implies $x=-1<y$ which contradicts our assumption so the only solution in this case is $x=\frac{5}{3}, y=z=-\frac{4}{3}$. Of course this gives rise to $6$ solutions by permuting the variables and introducing signs again.
Any other solutions for this system? I really think there is another method of solving (which may be elegant)
I think your method is probably best. The idea of fixing $x+y+z$ as non-negative and making $x$ the largest variable is an excellent method of reducing the number of cases.
However there are errors (typos?) in your solution. You can see this from your final solution, which contrary to your assumption actually has $x+y+z$ negative.
I wonder if you renumbered your equations when writing up your answer since right at the start you say "Subtracting the second from the first equation" when it's actually "Subtracting the third from the second equation". Later in the solution you have $y=x$ but then seem to switch to $y=z$.