Example 5 Let $p>2$ be a prime. Find the solutions to the congruence equation $x^{2} \equiv 1\left(\bmod p^{l}\right)$.
The congruence equation can be written as
$$(x-1)(x+1) \equiv 0\left(\bmod p^{l}\right)$$

Since $(x-1)+(x+1)=2$, the above equation is equivalent to
$$x-1 \equiv 0\left(\bmod p^{l}\right) \quad \text { or } \quad x+1 \equiv 0\left(\bmod p^{l}\right) \text {. }$$

Therefore, for any $l \geqslant 1$ the solutions are
$$x \equiv-1,1\left(\bmod p^{l}\right),$$

Solution 2.