When $y=1$, $7^y + 12y-1 = 18$. Therefore, we must have $x \le 18$. We are going to show that for any $y$, $18\mid (7^y + 12y-1)$. If so, the answer will be $\boxed{18}$.

Because $(7^y+12y-1)$ is obviously an even number, it is sufficient to show it is also a multiple of $9$ in order to prove it is a multiple of $18$.

• if $y\equiv 0\pmod{3}$, let it be $y=3k$ where $k$ is a positive integer. Then $$7^y+12y-1\equiv \left(7^3\right)^k + 12y-1\equiv 1^k+ 36k - 1\equiv 0\pmod{9}$$
• if $y\equiv 1\pmod{3}$, let it be $y=3k+1$ where $k$ is a positive integer. Then $$7^y+12y-1\equiv 7\cdot\left(7^3\right)^k + 12y-1\equiv 7\cdot 1^k+ 36k + 3 - 1\equiv 0\pmod{9}$$
• if $y\equiv 2\pmod{3}$, let it be $y=3k+2$ where $k$ is a positive integer. Then $$7^y+12y-1\equiv 7^2\cdot\left(7^3\right)^k + 12y-1\equiv 7^2\cdot 1^k+ 36k + 6 - 1\equiv 0\pmod{9}$$

Therefore, it is always a multiple of $9$.