Let us call a triple $(x, y, z)$ of real numbers a *-triple if it satisfies
$$x^2 + y^2 + z^2 + 2xyz = 1.$$
Let us call a polynomial $p(x)$ with real coefficients a *-polynomial if $(x, y, z)$ a *-triple implies $(p(x), p(y), p(z))$ a *-triple.
We first investigate polynomials of degree at most 1. Hence, assume that $p(x) = ax + b$ ($a, b \in \mathbb{R}$), and suppose that $p(x)$ is a *-polynomial. Since $(x, -x, 1)$ and $(x, x, -1)$ are *-triples for all $x \in \mathbb{R}$, we have
$$p(x)^2 + p(-x)^2 + p(1)^2 + 2p(x)p(-x)p(1) = 1 \text{ and } p(x)^2 + p(x)^2 + p(-1)^2 + 2p(x)p(x)p(-1) = 1$$
for all $x \in \mathbb{R}$. This simplifies, respectively, to
$$2a^2(1 - a - b)x^2 + 2b^2(1 + a + b) + (a + b)^2 = 1 \quad (1)$$
$$\text{and}\quad 2a^2(1 - a + b)x^2 + 4ab(1 - a + b)x + 2b^2(1 - a + b) + (b - a)^2 = 1. \quad (2)$$
Since these equations are valid for all $x \in \mathbb{R}$, we must have both the coefficients $2a^2(1-a-b)$ and $2a^2(1-a+b)$ equal to 0. Therefore, $a=0$, or $1-a-b=0=1-a+b$.
If $a = 0$, then, from (1), $2b^2(1+b) + b^2 = 1$. It is clear that $b = -1$ is a solution to this equation, and it follows readily that $b = \frac{1}{2}$ is the only other solution. It is straightforward to check that the constant polynomials $p(x) = -1$ and $p(x) = \frac{1}{2}$ are indeed *-polynomials.
In case $a \neq 0$, then $1 - a - b = 0 = 1 - a + b$, so that $b = 0$. From (1), the coefficient $2a^2(1 - a - b) = 2a^2(1 - a) = 0$, giving $a = 1$. From this we get the trivial *-polynomial $p(x) = x$.

Our next observation is that when $p(x)$ is a *-polynomial, then $p(p(x)) = p^2(x)$ is also a *-polynomial — and, in fact, it follows by an easy induction that $p^n(x)$ are *-polynomials for all $n \ge 1$, where $p^n(x)$ means the composition of $p(x)$ with itself, $n$ times. So if we can find a *-polynomial $p(x)$ such that infinitely many polynomials in the sequence $p(x), p^2(x), p^3(x), \dots$ are different from each other, the problem will be solved. Unfortunately, none of the three -polynomials we have found so far has this property. We therefore look for a possible second degree -polynomial $p(x) = ax^2 + bx + c$, $a, b, c \in \mathbb{R}$, $a \neq 0$. In this case, assuming that $p(x)$ is a -polynomial, and again using the -triples $(x, -x, 1)$ and $(x, x, -1)$ (for all $x \in \mathbb{R}$), we obtain, after simplification:
$$2a^2(1 + a + b + c)x^4 + 2[2ac + b^2 + (2ac - b^2)(a + b + c)]x^2 + 2c^2(1 + a + b + c) + (a + b + c)^2 = 1 \quad (3)$$
and
$$2a^2(1 + a - b + c)x^4 + 4ab(1 + a - b + c)x^3 + 2(2ac + b^2)(1 + a - b + c)x^2 + 4bc(1 + a - b + c)x + 2c^2(1 + a - b + c) + (a - b + c)^2 = 1 \quad (4)$$
Since the coefficients of $x^4$ in (3) and (4) must be 0, and we have $a \neq 0$, we must have $a + b + c = -1 = a - b + c$, so that $b = 0$. Thus $a + c = -1$ and we conclude that
$$p(x) = ax^2 + c = (-1-c)x^2 + c$$
for some $c \in \mathbb{R}$. But $p(x)$ is assumed to be a *-polynomial, and since $(x, \sqrt{1-x^2}, 0)$ are *-triples for all $-1 \le x \le 1$, we get that
$$((-1-c)x^2+c)^2 + ((-1-c)(1-x^2)+c)^2 + c^2 + 2((-1-c)x^2+c)((-1-c)(1-x^2)+c)c = 1,$$
which simplifies to
$$2(1+c)^2(1-c)x^4 - 2(1+c)^2(1-c)x^2 + 1 = 1.$$
As before, the coefficients of $x^4$ and $x^2$ must be 0, and we see that $c \in \{-1, 1\}$. The case $c = -1$ gives $p(x) = -1$, which we have already dealt with. Hence, the only possible candidate at this stage for a second degree *-polynomial, is $p(x) = -2x^2 + 1$. We now verify that $p(x) = -2x^2 + 1$ is indeed a *-polynomial:
Let $(x, y, z)$ be an arbitrary *-triple. Then
$$\begin{aligned} & (-2x^2 + 1)^2 + (-2y^2 + 1)^2 + (-2z^2 + 1)^2 + 2(-2x^2 + 1)(-2y^2 + 1)(-2z^2 + 1) \\ &= -16x^2y^2z^2 + 4(x^4 + y^4 + z^4) + 8(x^2y^2 + y^2z^2 + z^2x^2) - 8(x^2 + y^2 + z^2) + 5 \\ &= -4(1 - (x^2 + y^2 + z^2))^2 + 4(x^2 + y^2 + z^2)^2 - 8(x^2 + y^2 + z^2) + 5 \\ &= 1, \end{aligned}$$
and we conclude that $p(x) = -2x^2 + 1$ is indeed a *-polynomial.
This solves the problem, since we now have an infinite sequence $p(x), p^2(x), p^3(x), \dots$ of *-polynomials, and they are all different, since $\deg(p^n(x)) = 2^n$ for each $n \ge 1$, a fact that can easily be verified by induction.