Example 2 Find all natural numbers $n$ such that $13 \mid 7^{n}-1$.

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Analysis Using Fermat's Little Theorem, we know $13 \times 7^{12}-1$, and then proceed further.
Solution First, by Fermat's Little Theorem: $13 \mid 7^{12}-1$. Next, consider the case $1 \leqslant n<12$, and find that $13 \times 7^{n}-1$. Therefore, $13 \mid 7^{12 k}-1$, while $13 \nless 7^{12 k+r}-1,(1 \leqslant r<12)$. In fact, $7^{12 k+r}-1=7^{r}\left(7^{12 k}-1\right)+7^{r}-1$, so $13 \times 7^{12 k-r}-1,(1 \leqslant r<12)$. The condition is satisfied if and only if $n=12 k\left(k \in \mathbf{N}^{*}\right)$.