We distinguish several cases.

First, assume $x = y = z$. Then the whole system reduces to $(k+2)x^2 = x$. Its solution is a triplet $(0, 0, 0)$ for any $k$ and moreover triplet $\left(\frac{1}{k+2}, \frac{1}{k+2}, \frac{1}{k+2}\right)$ if $k \neq -2$.

Let's get back to the original system. Subtracting the second equation from the first one yields
$$(x^2 - z^2) + ky(x - z) = z - x$$
which rewrites as
$$(x - z)(x + z + ky + 1) = 0. \tag{1}$$
Similarly, subtracting the third equation from the second one yields
$$(y - x)(y + x + kz + 1) = 0. \tag{2}$$
If $x \neq y \neq z \neq x$, the equations (1), (2) reduce to
$$\begin{align*} x + z + ky + 1 &= 0, \\ y + x + kz + 1 &= 0. \end{align*}$$
Subtracting these two equations we arrive at $(y-z)(k-1) = 0$ implying that $k=1$
and $x+y+z=-1$. However that's impossible since for $k=1$ we get
$$z = x^2 + xy + y^2 = \left(x + \frac{y}{2}\right)^2 + \frac{3y^2}{4} \ge 0$$
and likewise $x \ge 0$ and $y \ge 0$ so altogether $x + y + z \ge 0$.

We found out that in every solution to the original system, some two unknowns
have the same value. As the system is cyclic, let us from now on assume $x = y \neq z$ (the
case $x = y = z$ has already been solved). Equation (1) then implies $x+y+ky+1=0$,
that is $x = -(k+1)y - 1$, and the original system reduces to a single equation
$$(k + 2)y^2 + (k + 1)y + 1 = 0. \tag{3}$$
Let us remark that any solution to equation (3) is a solution we haven't found yet, because equality $x = y$ i.e. $y = -(k+1)y - 1$ is only possible for $k \neq -2$ and yields $x = y = z = -1/(k+2)$ which is not a solution to the original system.

For $k = -2$ the equation (3) is linear with a unique solution $y = 1$. This yields
solution $(0, 1, 1)$ and its two permutations.

For $k \neq -2$ the equation (3) is quadratic and has real solutions if and only if
$$D = (k + 1)^2 - 4(k + 2) = k^2 - 2k - 7 \geq 0,$$
which translates to $k \notin (1-2\sqrt{2}, 1+2\sqrt{2})$. For $k = 1 \pm 2\sqrt{2}$ there is a unique solution
$$y_0 = -\frac{k+1}{2(k+2)} = 1 \mp \sqrt{2} \quad \text{a} \quad x_0 = \frac{(k+1)^2}{2(k+2)} - 1 = 1.$$
which yields the three permutations of $(x_0, y_0, y_0)$ as solutions to the original system.

For $k \in (-\infty, -2) \cup (-2, 1-2\sqrt{2}) \cup (1+2\sqrt{2}, \infty)$, the quadratic equation (3) has
two distinct solutions
$$y_{1,2} = \frac{-k - 1 \pm \sqrt{k^2 - 2k - 7}}{2(k + 2)},$$
that give two distinct values $x_{1,2} = -(k+1)y_{1,2} - 1$. The original system thus has six solutions: three permutations of $(x_1, y_1, y_1)$ and three permutations of $(x_2, y_2, y_2)$.

The following table summarizes the number of solutions to the given system in
terms of $k$:

| Interval for $k$ | $(0, 0, 0)$ | $(1/(k+2), 1/(k+2), 1/(k+2))$ | Equation (3) | Total |
|------------------|-------------|-------------------------------|-------------|-------|
| $(-\infty, -2)$ | 1 | 1 | 6 | 8 |
| $-2$ | 1 | 0 | 3 | 4 |
| $(-2, 1 - 2\sqrt{2})$ | 1 | 1 | 6 | 8 |
| $1 - 2\sqrt{2}$ | 1 | 1 | 3 | 5 |
| $(1 - 2\sqrt{2}, 1 + 2\sqrt{2})$ | 1 | 1 | 0 | 2 |
| $1 + 2\sqrt{2}$ | 1 | 1 | 3 | 5 |
| $(1 + 2\sqrt{2}, \infty)$ | 1 | 1 | 6 | 8 |