Example 1 Let $n$ be a given positive integer. Find all positive numbers $a, b$ such that $x^{2}+a x+b$ is a factor of $a x^{2 n}+(a x+b)^{2 n}$.


Let $x_{0}$ be a root of $x^{2} + a x + b = 0$.
Given $b > 0$, we know $x_{0} \neq 0$, and $x_{0}$ is also a root of
$$
a x^{2 n} + (a x + b)^{2 n} = 0
$$

Thus,
$$
\begin{array}{l}
\text { Hence } a x_{0}^{2 n} + (a x_{0} + b)^{2 n} = a x_{0}^{2 n} + (-x_{0}^{2})^{2 n} \\
= a x_{0}^{2 n} + x_{0}^{4 n} = 0 \\
\Rightarrow a + x_{0}^{2 n} = 0 \\
\Rightarrow x_{0}^{2 n} = -a = a(\cos \pi + \mathrm{i} \sin \pi) \\
\Rightarrow x_{0} = \sqrt[2 n]{a} \left[\cos \frac{(2 k + 1) \pi}{2 n} + \mathrm{i} \sin \frac{(2 k + 1) \pi}{2 n}\right],
\end{array}
$$

where $k = 0, 1, \cdots, 2 n - 1$.
On the other hand, by the quadratic formula, we have
$$
\begin{array}{l}
x_{0} = \frac{-a \pm \sqrt{4 b - a^{2}} \mathrm{i}}{2}. \\
\text { Hence } \sqrt{2 n} \cos \frac{(2 k + 1) \pi}{2 n} = -\frac{a}{2} \\
\Rightarrow \cos \frac{(2 k + 1) \pi}{2 n} < 0 \\
\Rightarrow n < 2 k + 1 < 3 n.
\end{array}
$$

Therefore, when $n = 1$, there is no solution;
when $n \geqslant 2$,
$$
\begin{array}{l}
a = \left[-2 \cos \frac{(2 k + 1) \pi}{2 n}\right]^{\frac{2 n}{2 n - 1}}, \\
b = x_{0} \overline{x_{0}} = |x_{0}|^{2} = \sqrt[n]{a} \\
= \left[2 \cos \frac{(2 k + 1) \pi}{2 n}\right]^{\frac{2}{2 n - 1}},
\end{array}
$$

where $n < 2 k + 1 < 3 n, k \in \mathbf{Z}_{+}$.