The roots of the polynomial
$$ f(x)= x^8 + 8 x^6 + 60 x^4 + 176 x^2 + 2187 x + 1942 $$
allow to satisfy the conditions for $x \ne y \ne z$.
We rediscover the 2 solutions $x=y=z=-1$ and $x=y=z=-2$ but also one complex solution
$(x,y,z)=(2.44944 -1.66645i,-1.74091+2.72124i, 0.791465+3.15828i) $
where the rotations $(y,z,x), (z,x,y)$ and the complex conjugates $(\overline x,\overline y, \overline z)$ are also solutions.
If we write $$g(x) = -{x^2+2\over 3}$$
then your set of equations is also
$$ \begin{eqnarray} y=g(x)&;& z=g(y)=g(g(x))&;& x=g(z)=g(g(g(x))) \end{eqnarray}$$
and the equation $x = g(g(g(x))))$ leads to the polynomial $f(x)$ above.
I've computed the roots using Pari/GP and get them to the first few digits:
$$ \small
\begin{array} {rr}
-2.00000000000 \\
-1.00000000000 \\
2.44944037937&-1.66644590342*I \\
2.44944037937&+1.66644590342*I \\
-1.74090540769&-2.72123992391*I \\
-1.74090540769&+2.72123992391*I \\
0.791465028319&-3.15828086610*I \\
0.791465028319&+3.15828086610*I
\end{array}$$
We rewrite the set of equations
$$ \begin{eqnarray} y = -{x^2+2\over3} & ;&z = -{y^2+2\over3}&;&x = -{z^2+2\over3} \end{eqnarray}$$
If I insert the roots I get for the successive evalutations the following ( and the last entry must equal the first for each row):
$$ \small
\begin{array} {rr|rr|rr|rr}
x=root_r&& y=g(x) && z=g(y) && x=g(z) \\
\hline
-2.00000 && -2.00000 && -2.00000 && -2.00000 \\
-1.00000 && -1.00000 && -1.00000 && -1.00000 \\
2.44944 &-1.66645i & -1.74091&+2.72124i & 0.791465&+3.15828i & 2.44944&-1.66645i \\
2.44944 &+1.66645i & -1.74091&-2.72124i & 0.791465&-3.15828i & 2.44944&+1.66645i \\
-1.74091&-2.72124i & 0.791465&-3.15828i & 2.44944&+1.66645i & -1.74091&-2.72124i \\
-1.74091&+2.72124i & 0.791465&+3.15828i & 2.44944&-1.66645i & -1.74091&+2.72124i \\
0.791465&-3.15828i & 2.44944&+1.66645i & -1.74091&-2.72124i & 0.791465&-3.15828i \\
0.791465&+3.15828i & 2.44944&-1.66645i & -1.74091&+2.72124i & 0.791465&+3.15828i
\end{array} $$
We can express the 8 possible solutions of sets $(x,y,z)$ by the following indexes into the polroots, where we begin counting at 1:
$[1,1,1],[2,2,2],[3,6,8],[4,5,7],[5,7,4],[6,8,3],[7,4,5],[8,3,6]$
The first two solutions have $x=y=z$
while the following 6 solutions have $x \ne y \ne z$ (being rotations and complex conjugates of each other)