Example 3 Simplify the congruence equation
$$21 x^{18}+2 x^{15}-x^{10}+4 x-3 \equiv 0(\bmod 7)$$
Solve: First, remove the terms with coefficients that are multiples of 7 to get
$$2 x^{15}-x^{10}+4 x-3 \equiv 0(\bmod 7)$$

Perform polynomial division (19) $(p=7)$ to get
$$\begin{aligned}
2 x^{15}-x^{10}+4 x-3= & \left(x^{7}-x\right)\left(2 x^{8}-x^{3}+2 x^{2}\right) \\
& +\left(-x^{4}+2 x^{3}+4 x-3\right)
\end{aligned}$$

This yields the equivalent congruence equation
$$x^{4}-2 x^{3}-4 x+3 \equiv 0(\bmod 7)$$

By directly substituting $x=0, \pm 1, \pm 2, \pm 3$, we find that the congruence equation has no solutions.