jmc

We write $12$ and $18$ as products of $2$s and $3$s: $$\begin{array}{rl}12^2\cdot 18^3& =(2^2\cdot 3{)}^2\cdot (2\cdot 3^2{)}^3\\ & =(2^4\cdot 3^2)\cdot (2^3\cdot 3^6)\\ & =2^{4+3}\cdot 3^{2+6}\\ & =2^7\cdot 3^8\end{array}$$Therefore, $x+y=7+8=\boxed{15}$.