Example 9 Find all positive integers that are coprime with all terms of the sequence $\left\{a_{n}\right\}$ satisfying
$$
a_{n}=2^{n}+3^{n}+6^{n}-1\left(n \in Z_{+}\right)
$$
Solution: Clearly, $\left(1, a_{n}\right)=1$.
Let $m(m>1)$ be a positive integer that is coprime with all terms in $\left\{a_{n}\right\}$, and let $p$ be a prime factor of $m$.
If $p>3$, then by Fermat's Little Theorem,
$$
\begin{array}{l}
2^{p-1} \equiv 1(\bmod p), 3^{p-1} \equiv 1(\bmod p), \\
6^{p-1} \equiv 1(\bmod p) .
\end{array}
$$

Let $2^{p-1}=m p+1,3^{p-1}=n p+1$,
$$
6^{p-1}=t p+1\left(m, n, t \in \mathbf{Z}_{+}\right) \text {. }
$$

Then $a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1$
$$
\begin{array}{l}
=\frac{m p+1}{2}+\frac{n p+1}{3}+\frac{t p+1}{6}-1 \\
=\frac{3 m p+2 n p+t p}{6}=p \cdot \frac{3 m+2 n+t}{6} .
\end{array}
$$

Since $a_{p-2}$ is an integer, $(p, 6)=1$, thus,
$$
6|(3 m+2 n+t), p| a_{p-2} \text {. }
$$

This contradicts $\left(m, a_{p-2}\right)=1$.
If $p=2$ or 3, then $a_{2}=48=2^{4} \times 3 \Rightarrow p \mid a_{2}$, which is also a contradiction.

Therefore, the only positive integer that is coprime with all terms in the sequence $\left\{a_{n}\right\}$ is 1.