Example 1 (24th All-Soviet Union Mathematical Olympiad, 1990) Determine for which natural numbers $n$, the number $3^{2 n+1}-2^{2 n+1}-6^{n}$ is composite.

Determine for which natural numbers $n$, the number $3^{2 n+1}-2^{2 n+1}-6^{n}$ is composite.
The original expression can be transformed as follows:
$$
\begin{aligned}
3^{2 n+1}-2^{2 n+1}-6^{n} & =3^{2 n+1}-3^{n} \cdot 2^{n}-2^{2 n+1} \\
& =3^{2 n+1}-3^{n+1} \cdot 2^{n}+3^{n} \cdot 2^{n+1}-2^{2 n+1} \\
& =3^{n+1}\left(3^{n}-2^{n}\right)+2^{n+1}\left(3^{n}-2^{n}\right) \\
& =\left(3^{n}-2^{n}\right)\left(3^{n+1}+2^{n+1}\right) .
\end{aligned}
$$

When $n=1$, $3^{2 n+1}-2^{2 n+1}-6^{n}=13$ is a prime number.
When $n>1$, since
$$
3^{n}-2^{n}>1,3^{n+1}+2^{n+1}>1 \text {, }
$$

$3^{2 n+1}-2^{2 n+1}-6^{n}$ is a composite number.
Therefore, the required $n$ is a natural number where $n \geqslant 2$.