Example 4 Find all prime numbers $p$ such that $p$ is coprime with every term of the sequence given by the general term formula
$$
a_{n}=20^{n}+15^{n}+12^{n}-10^{n}\left(n \in \mathbf{Z}_{+}\right)
$$
【Analysis】It can be verified that $p=2,5$ satisfy the problem.
Consider
$$
a_{2}=12^{2}+15^{2}+20^{2}-10^{2}=12^{2}+15^{2}+300 \text {, }
$$
$3 \mid a_{2}$.
Obviously, $p \neq 3$.
We will prove below: When $p \geqslant 7$, $p$ must divide some term in the sequence $\left\{a_{n}\right\}$.
First, we find such a term by expanding the definition: $a_{m}=0$.
Let $m=-k(k>0)$. Then
$$
\begin{array}{l}
20^{-k}+15^{-k}+12^{-k}-10^{-k} \\
=\frac{3^{k}+4^{k}+5^{k}-6^{k}}{60^{k}}=0 \\
\Rightarrow 3^{k}+4^{k}+5^{k}-6^{k}=0 .
\end{array}
$$

By the monotonicity of $f(x)=\left(\frac{3}{6}\right)^{x}+\left(\frac{4}{6}\right)^{x}+\left(\frac{5}{6}\right)^{x}-1$, we know
Equation (1) $\Leftrightarrow k=3$.
Consider $a_{-3}=0$.
Following Example 3, we know that when $p \geqslant 7$, $p \mid a_{p-4}$.
In summary, the prime numbers that satisfy the problem are $2$ and $5$.