Example 6 Let $n$ be a given natural number. Find all positive pairs $(a, b)$ such that $x^{2}+a x+b$ is a factor of $a x^{2 n}+$ $(a x+b)^{2 n}$.
(1992, Shanghai High School Mathematics Competition)
Let $x_{0}$ be a root of the equation $x^{2} + a x + b = 0$. From $b > 0$, we know $x_{0} \neq 0$, and $x_{0}$ is also a root of the equation
$$
a x^{2 n} + (a x + b)^{2 n} = 0
$$

Thus, $a x_{0}^{2 n} + \left(-x_{0}^{2}\right)^{2 n} = 0$, which gives $x_{0}^{2 n} = -a$. Solving this, we get
$$
x_{0} = \sqrt[2 n]{a} \left[\cos \frac{(2 k + 1) \pi}{2 n} + \mathrm{i} \sin \frac{(2 k + 1) \pi}{2 n}\right]
$$
$(k = 0, 1, 2, \cdots, 2 n - 1)$, and $x_{0}$ must be a complex number, otherwise $-a = x_{0}^{2 n} \geqslant 0$, which is a contradiction.
On the other hand, $x_{0} = \frac{-a \pm \sqrt{4 b - a^{2}} \cdot \mathrm{i}}{2}$, so
$$
\sqrt[2 n]{a} \cdot \cos \frac{(2 k + 1) \pi}{2 n} = -\frac{a}{2} < 0.
$$

Thus, $\frac{\pi}{2} < \frac{(2 k + 1) \pi}{2 n} < \frac{3 \pi}{2}$.
Hence, $n < 2 k + 1 < 3 n$.
From this, we see that when $n = 1$, there is no solution;
When $n \geqslant 2$,
$$
\begin{array}{l}
a = \left[2 \cos \frac{(2 k + 1) \pi}{2 n}\right]^{\frac{2 n}{2 n - 1}}, \\
b = x_{0} \overline{x_{0}} = \left|x_{0}\right|^{2} = a^{\frac{1}{n}} = \left[2 \cos \frac{(2 k + 1) \pi}{2 n}\right]^{\frac{2}{2 n - 1}},
\end{array}
$$

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