A cyclic system

Pairwise subtraction. Subtracting the second equation from the first:
$$x^2-y^2=z(y-x)\ \Longleftrightarrow\ (x-y)(x+y+z)=0.$$
Similarly,
$$(y-z)(x+y+z)=0,\qquad (z-x)(x+y+z)=0.$$

Case 1: $x+y+z\ne0$. Then all three factors in the first position must vanish, so $x=y=z$. The first equation gives $x^2=x^2+1$, impossible. No solutions.

Case 2: $x+y+z=0$. Then the three subtractions give no further constraint. Substitute $z=-(x+y)$ in the first equation:
$$x^2=-y(x+y)+1\quad\Longleftrightarrow\quad x^2+xy+y^2=1.$$
Cyclically, $y^2+yz+z^2=1$ and $z^2+zx+x^2=1$; with $x+y+z=0$ these are all the same equation. Adding the three original equations,
$$x^2+y^2+z^2=xy+yz+zx+3.$$
Using $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=0$, we get $xy+yz+zx=-\tfrac12(x^2+y^2+z^2)$, so
$$x^2+y^2+z^2=-\tfrac12(x^2+y^2+z^2)+3\ \Longrightarrow\ x^2+y^2+z^2=2.$$
So the solution set is contained in $\{x+y+z=0,\ x^2+y^2+z^2=2\}$, a circle of radius $\sqrt2$ in the plane $x+y+z=0$.

Additional constraint from $x^2=yz+1$. With $y+z=-x$ and $yz=x^2-1$, the numbers $y,z$ are the roots of
$$t^2+xt+(x^2-1)=0.$$
Its discriminant is $x^2-4(x^2-1)=4-3x^2$, so real $y,z$ require $|x|\le 2/\sqrt3$. By symmetry, $|y|\le 2/\sqrt3$ and $|z|\le 2/\sqrt3$.

Conclusion. The solution set is
$$\Bigl\{(x,y,z)\in\mathbb{R}^3\ :\ x+y+z=0,\ \ x^2+y^2+z^2=2,\ \ |x|\le 2/\sqrt3,\ |y|\le 2/\sqrt3,\ |z|\le 2/\sqrt3\Bigr\}.$$
On the circle $\{x+y+z=0,\ x^2+y^2+z^2=2\}$ the maximum of $|x|$ is $\sqrt2>2/\sqrt3$, so the inequality constraints cut the circle into three arcs: on each arc, one of $|x|,|y|,|z|$ reaches its largest value and the other two are equal. The boundary points of these arcs are the six permutations of
$$\Bigl(\tfrac1{\sqrt3},\,\tfrac1{\sqrt3},\,-\tfrac2{\sqrt3}\Bigr)\quad\text{and}\quad\Bigl(-\tfrac1{\sqrt3},\,-\tfrac1{\sqrt3},\,\tfrac2{\sqrt3}\Bigr).$$
In particular, the solution set is a continuum (three closed arcs), not a discrete set; the six permutations are its boundary points, not additional solutions.

Check of a boundary point. For $(x,y,z)=\bigl(\tfrac1{\sqrt3},\tfrac1{\sqrt3},-\tfrac2{\sqrt3}\bigr)$: $x^2=\tfrac13$ and $yz+1=-\tfrac23+1=\tfrac13$ ✓; $y^2=\tfrac13$ and $zx+1=-\tfrac23+1=\tfrac13$ ✓; $z^2=\tfrac43$ and $xy+1=\tfrac13+1=\tfrac43$ ✓.