Example 2 Solve the congruence equation $x^{12} \equiv 17\left(\bmod 2^{6}\right)$.
From Table 4 in Example 2 of §3, we find that the indicator set of 17 is $\gamma^{(-1)}(17)=0, \gamma^{(0)}(17)=12$. Therefore, we need to solve two linear congruences:
$$\begin{array}{ll}
12 u \equiv 0(\bmod 2), & 0 \leqslant u<1 \\
12 v \equiv 12\left(\bmod 2^{4}\right), & 0 \leqslant v<2^{4}
\end{array}$$

It is easy to see that $u=0,1 ; v=1,5,9,13$. Thus, by formula (11), we obtain eight solutions for $x$. Referring to Table 3 in Example 2 of §3, these eight solutions are
$$x \equiv 5,-11,-27,21,-5,11,27,-21\left(\bmod 2^{6}\right) \text {. }$$