Example 2 Solve the congruence equation $x^{3}+5 x^{2}+9 \equiv 0\left(\bmod 7 \cdot 3^{4}\right)$.
Solve: By Theorem 1, we need to solve the system of congruences
$$\left\{\begin{array}{l}
x^{3}+5 x^{2}+9 \equiv 0(\bmod 7), \\
x^{3}+5 x^{2}+9 \equiv 0\left(\bmod 3^{4}\right)
\end{array}\right.$$

By direct calculation, the solution to the first congruence is
$$x \equiv-2(\bmod 7)$$

By Example 1, the solution to the second congruence is
$$x \equiv-21,6,33,-24,3,30,40\left(\bmod 3^{4}\right) .$$

Then, solve the system of linear congruences
$$\left\{\begin{array}{l}
x \equiv a_{1}(\bmod 7) \\
x \equiv a_{2}\left(\bmod 3^{4}\right)
\end{array}\right.$$

Using Theorem 1 in § 3, here $m_{1}=M_{2}=7, m_{2}=M_{1}=3^{4}$. By
$$M_{1} \equiv 9^{2} \equiv 2^{2} \equiv-3(\bmod 7)$$

we can take $M_{1}^{-1}=2$. By $M_{2} \equiv 7\left(\bmod 3^{4}\right)$, we can take $M_{2}^{-1}=-23$. Therefore, the solution to this system of linear congruences is
$$\begin{aligned}
x & \equiv 3^{4} \cdot 2 \cdot a_{1}+7 \cdot(-23) \cdot a_{2} \\
& \equiv 162 a_{1}-161 a_{2}(\bmod 567) .
\end{aligned}$$

Substituting $a_{1}=-2, a_{2}=-21,6,33,-24,3,30,40$ respectively, we get $x \equiv$ $3057,-1290,-5637,3540,-807,-5154,-676(\bmod 567)$, i.e.,
$$x \equiv 222,-156,33,138,-240,-51,40(\bmod 567)$$

These are all the solutions to the required congruence equation, with 7 solutions.