Example 5 Find the largest positive integer $x$, such that for every positive integer $y$, we have $x \mid\left(7^{y}+12 y-1\right)$.


When $y=1$, $7^{y}+12 y-1=18$.
Let 18 I $\left(7^{y}+12 y-1\right)$. Notice that,
$$
\begin{array}{l}
7^{y+1}+12(y+1)-1 \\
=6 \times\left(7^{y}+2\right)+\left(7^{y}+12 y-1\right) .
\end{array}
$$

Since $7^{y}+2 \equiv 1+2 \equiv 0(\bmod 3)$, therefore,
$$
18 \mid\left[7^{y+1}+12(y+1)-1\right] \text {. }
$$

Thus, for any positive integer $y, 18 \mid\left(7^{y}+12 y-1\right)$.
Hence, the required maximum positive integer $x=18$.