Example 8 Try to solve the congruence equation
$$x^{2} \equiv 33(\bmod 128)$$
We have $1^{2}-33=-32$. So 1 is a root modulo 32, therefore, $1+16=17$ is a root modulo 64, because
$$17^{2}-33=289-33=256,$$

Hence 17 is also a root modulo 128, and the other root modulo 128 is
$$17+64=81 \equiv-47(\bmod 128)$$

Thus, the roots of the original congruence equation are
$$x \equiv \pm 17, \pm 47(\bmod 128)$$