Example 9 Try to solve the congruence equation
$$x^{2} \equiv 105(\bmod 256)$$
From $1^{2}-105=-104=-8 \times 13$ we know that 1 is not a root modulo 16, so 5 is a root modulo 16. Then, by
$$5^{2}-105=-80=-16 \times 5$$

we know that 5 is not a root modulo 32, hence $5+8=13$ is a root modulo 32. And by
$$13^{2}-105=169-105=64$$

we know that 13 is not only a root modulo 32 but also a root modulo 64, thus $13+32=45$ is a root modulo 128, because
$$45^{2}-105=2025-105=1920=128 \times 15,$$

so 45 is not a root modulo 256, hence $45+64=109$ is a root modulo 256, and the other root modulo 256 is
$$109+128=237 \equiv-19(\bmod 256)$$

Therefore, the roots of the original congruence equation are
$$x \equiv \pm 19, \pm 109(\bmod 256)$$