Naturally

Because the multiplier $12 = 2^2\cdot 3$ contains 2 and 3 as prime factors, the smallest values for $x$ and $y$ will only contain these prime factors. So we write $x = 2^a\cdot 3^b, \ \ \ \ \ y = 2^c\cdot 3^d.$ The equation becomes $(2^a\cdot 3^b)^5 = 2^2\cdot 3\cdot (2^c\cdot 3^d)^8.$ Distributing the powers and comparing the exponents of 2 and 3, we get the linear equations $\left\{\begin{array}{l} 5a = 2+8c; \\ 5b = 1 + 8d \end{array}\right.$ We must find the smallest non-negative integer solution for each of these equations. It is not difficult to see that $\left\{\begin{array}{l} 5\cdot 2 = 2+8\cdot 1; \\ 5\cdot 5 = 1 + 8\cdot 3 \end{array}\right.$

Thus we have $a = 2, b = 1, c = 5, d = 3$ and $x = 2^2\cdot 3^5 = 972;\ \ \ \ y = 2^1\cdot 3^2 = 54,$ so that the answer is $\boxed{1026}$ .

Note: The Chinese remainder theorem guarantees that there is a unique solution with $a, c < 8$ and $b, d < 5$ .

All other solutions are of the form $x = 972\cdot n^8,\ \ \ \ y = 54\cdot n^5,$ where $n$ is any positive integer. (Prove this!)