Example 5 Find all positive integers that are coprime with all terms of the sequence $\left\{a_{n}=2^{n}+3^{n}+6^{n}-1, n \geqslant\right.$ $1\}$.
Solution: Clearly, $\left(1, a_{n}\right)=1$. Let $m>1$ 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, we have:
$$
2^{p-1} \equiv 1(\bmod p), 3^{p-1} \equiv 1(\bmod p), 6^{p-1} \equiv
$$
$1(\bmod p)$.
$$
\text { Let } 2^{p-1}=m p+1,3^{p-1}=n p+1,6^{p-1}=t p
$$
$+1(m, n, t \in \mathbf{Z})$, then
$$
\begin{array}{l}
a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1=\frac{m p+1}{2}+ \\
\frac{n p+1}{3}+\frac{t p+1}{6}-1 \\
\quad=\frac{3 m p+2 n p+t p}{6}=p \cdot \frac{3 m+2 n+t}{6} . \\
\because \text { Since } a_{p-2} \text { is an integer, }(p, 6)=1, \\
\therefore 6|3 m+2 n+t, p| a_{p-2}, \text { which contradicts }\left(m, a_{p-2}\right)=
\end{array}
$$
$\because$ Since $a_{p-2}$ is an integer, $(p, 6)=1$,
1, contradiction! If $p=2$ or 3, then $a_{2}=48=2^{4} \cdot 3, p \mid$ $a_{2}$, also a contradiction! Therefore, the only positive integer that is coprime with all terms in $\left\{a_{n}\right\}$ is 1.