Example 8 Find the largest positive integer $x$, such that for every positive integer $y$, $x$ divides $7^{y}+12 y-1$.

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Analysis Since $7^{x}+12 y-1$ increases with the increase of $y$, when $y=1$, it takes the minimum value 18, therefore, from $x \mid (7^{y}+12 y-1)$, we get $x \mid 18$, so $x$ is a divisor of 18. Also, when $y=2$, $7^{2}+12 \times 2-1=72=18 \times 4$, and when $y=3$, $7^{3}+12 \times 3-1=378=18 \times 21$, both have $18 \mid 7^{y}+12 y-1$, hence we guess that the $x$ that meets the condition is $x=18$.

Solution Using mathematical induction, when $y=1$, the guess holds. Inductive hypothesis: when $y=k$, the guess is true, i.e., $18 \mid (7^{k}+12 k-1)$, then when $y=k+1$,
$$
7^{k+1}+12(k+1)-1=7^{k}+12 k-1+6 \times 7^{k}+12=(7^{k}+12 k-1)+6(7^{k}+2).
$$

By the inductive hypothesis, we know that $7^{k}+12 k-1$ can be divided by 18.
Therefore, we only need to prove that $7^{k}+2$ can be divided by 3.
In fact, $7^{k}=(6+1)^{k}$
$$
\begin{array}{l}
=6^{k}+C_{k}^{1} \cdot 6^{k-1}+C_{k}^{2} \cdot 6^{k-2}+\cdots+C_{k}^{k} \\
=6(6^{k-1}+C_{k}^{1} \cdot 6^{k-2}+C_{k}^{2} \cdot 6^{k-3}+\cdots+C_{k}^{k-1})+1,
\end{array}
$$

Thus, $7^{k}$ leaves a remainder of 1 when divided by 3. Therefore, $7^{k}+2$ can be divided by 3.
Therefore, when $y=k+1$, $18 \mid (7^{k+1}+12(k+1)-1)$, thus proving the problem by induction.