Example 4 (1992 "Friendship Cup" International Mathematics Competition Question) Find the largest natural number $x$, such that for every natural number $y$, $x$ divides $7^{y}+12 y-1$.
When $y=1$, $7^{y}+12 y-1=18$, therefore $x \mid 18$, thus $x \leqslant 18$.
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The following is a proof by mathematical induction:
$18 \mid 7^{y}+12 y-1$.
When $y=1$, it is clearly true.
If $y=k$, then $18 \mid 7^{k}+12 k-1$.
When $y=k+1$,
\[
\begin{aligned}
& 7^{k+1}+12(k+1)-1 \\
= & 7 \cdot 7^{k}+7 \cdot 12 k-7-6 \cdot 12 k+18 \\
= & 7\left(7^{k}+12 k-1\right)-72 k+18 .
\end{aligned}
\]

By the induction hypothesis,
\[
18 \mid 7^{k}+12 k-1,
\]

and $18 \mid -72 k+18$,
thus $18 \mid 7^{k+1}+12(k+1)-1$,
i.e., the proposition holds for $y=k+1$.
Therefore, for all natural numbers $y$, $18 \mid 7^{y}+12 y-1$.
Thus, the largest natural number $x=18$.