Let $\overline{x 6 y 4}=A^{2}$, then $A$ is a two-digit number greater than 40, and its unit digit is 2 or 8. It is not difficult to verify that only when $A=68$ or 98, the hundred's digit of $A^{2}$ is 6. Since $68^{2}=4624,98^{2}=9604$, $(x, y)$ can only be $(4,2)$ and $(9,0)$. Also,
$$
\begin{aligned}
\underbrace{44 \cdots 4}_{n \uparrow} 6 \underbrace{22 \cdots 24}_{n \uparrow} & =\frac{1}{9}\left(4 \cdot 10^{2(n+1)}+16 \cdot 10^{n+1}+16\right) \\
& =\left[\left(2 \cdot 10^{n+1}+4\right) / 3\right]^{2}=\underbrace{66 \cdots 6}_{n \uparrow} 8^{2} . \\
\underbrace{99 \cdots 9}_{n \uparrow} \underbrace{00 \cdots 04}_{n \uparrow} & =\left(10^{n}-1\right) \cdot 10^{n+2}+6 \cdot 10^{n+1}+4 \\
& =\left(10^{n+1}-2\right)^{2}=\underbrace{99 \cdots 9}_{n \uparrow} 8^{2} .
\end{aligned}
$$

Therefore, $x=4, y=2$ or $x=9, y=0$.
15 numbers $x(\neq 0)$ and $y$ such that for any $n \geqslant 1$, the number $\underbrace{x x \cdots x 6}_{n \uparrow} \underbrace{y y \cdots y}_{n \uparrow}$ is a perfect square. Find $x$ and $y$.