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.

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.