II. (40 points) Find all real-coefficient polynomials $P(x)$ such that $P(x)$ is non-constant and
$$
P\left(x^{2}-2 x+4\right) \mid P\left(x^{3}+8\right) \text {. }
$$
II. $P(x)=a x^{n}\left(a \in \mathbf{R}, a \neq 0, n \in \mathbf{Z}_{+}\right)$.
Below we prove that $P(x)=0$ has only the unique root $x=0$.
Assume there exists $\left|\alpha_{1}\right|>0$, such that $P\left(\alpha_{1}\right)=0$.
Let $\beta_{1} 、 \beta_{2}$ be the roots of $x^{2}-2 x+4=\alpha_{1}$. Then $\beta_{1}+\beta_{2}=2$.
By the problem statement, there exists a polynomial $Q(x)$ such that $P\left(x^{3}+8\right)=P\left(x^{2}-2 x+4\right) Q(x)$.
Then $P\left(\beta_{1}^{3}+8\right)=P\left(\beta_{1}^{2}-2 \beta_{1}+4\right) Q\left(\beta_{1}\right)$ $=P\left(\alpha_{1}\right) Q(x)=0$.
Similarly, $P\left(\beta_{2}^{3}+8\right)=0$.
$$
\begin{array}{l}
\text { By } \beta_{1}^{3}+8+\beta_{2}^{3}+8=\alpha_{1}\left(\beta_{1}+2\right)+\alpha_{1}\left(\beta_{2}+2\right) \\
=\alpha_{1}\left(\beta_{1}+\beta_{2}+4\right)=6 \alpha_{1},
\end{array}
$$

we know
$$
\begin{array}{l}
\left|\beta_{1}^{3}+8\right|+\left|\beta_{2}^{3}+8\right| \geqslant\left|\beta_{1}^{3}+8+\beta_{2}^{3}+8\right| \\
=\alpha_{1}\left(\beta_{1}+2\right)+\alpha_{1}\left(\beta_{2}+2\right)=6\left|\alpha_{1}\right| .
\end{array}
$$

Thus, $\max \left\{\left|\beta_{1}^{3}+8\right|,\left|\beta_{2}^{3}+8\right|\right\} \geqslant 3\left|\alpha_{1}\right|$.
Without loss of generality, assume $\left|\beta_{1}^{3}+8\right| \geqslant 3\left|\alpha_{1}\right|$.
Let $\alpha_{2}=\beta_{1}^{3}+8$, then
$\left|\alpha_{2}\right| \geqslant 3\left|\alpha_{1}\right|>\left|\alpha_{1}\right|$.
Since $\alpha_{2}$ is a root of $P(x)=0$, we can replace $\alpha_{1}$ with $\alpha_{2}$.

Repeating the above steps yields infinitely many distinct roots of $P(x)=0$, which is impossible.
Therefore, $P(x)=0$ has only the unique root $x=0$, i.e., $P(x)=a x^{n}\left(a \in \mathbf{R}, a \neq 0, n \in \mathbf{Z}_{+}\right)$.