Example 3 Determine all polynomials $p(x)$ satisfying:
$$
p\left(x^{2}+1\right)=[p(x)]^{2}+1, \quad p(0)=0.
$$
(32nd Putnam Mathematical Competition $\mathrm{A}-2$)
Solution: From the given conditions, we have
$$
\begin{array}{l}
p(0)=0, \\
p(1)=[p(0)]^{2}+1=1, \\
p(2)=[p(1)]^{2}+1=2, \\
p(5)=[p(2)]^{2}+1=5, \\
p\left(5^{2}+1\right)=[p(5)]^{2}+1=26, \\
\cdots \cdots,
\end{array}
$$

That is, $p(x)$ has infinitely many fixed points $0,1,2,5,26, \cdots$.
Since $p(x)$ is a polynomial, the equation $p(x)-x=0$ is an algebraic equation. And a univariate $n$-degree algebraic equation has only $n$ roots, so,
$p(x)-x \equiv 0$, i.e., $p(x)=x$.
On the other hand, $p(x)=x$ satisfies $p(0)=0$ and $p\left(x^{2}+1\right)=[p(x)]^{2}+1$, hence $p(x)=x$ is the unique polynomial sought.