Problem 1548

[Solution] Let $x_{0}, y_{0}$ be integers that satisfy the equation. Then
$$\underbrace{\sqrt{x_{0}+\sqrt{x_{0}+\sqrt{x_{0}+\cdots+\sqrt{x_{0}}}}}}_{1964 \text { times }}=y_{0}$$

Squaring and rearranging terms, we get
$$\underbrace{\sqrt{x_{0}+\sqrt{x_{0}+\cdots+\sqrt{x_{0}}}}}_{1963 \text { times }}=y_{0}^{2}-x_{0}.$$

That is, $\underbrace{\sqrt{x_{0}+\sqrt{x_{0}+\cdots+\sqrt{x_{0}}}}}_{1963 \text { times }}=m_{1}$,
where $m_{1}=y_{0}^{2}-x_{0}$, which is an integer.
Repeating the process of squaring and rearranging terms, we get
$$\sqrt{x_{0}+\sqrt{x_{0}}}=m$$

and $\sqrt{x_{0}}=k$, where $m$ and $k$ are both integers.
Thus, we have $x_{0}=k^{2}$,
which gives
$$\begin{array}{l} k^{2}+k=m^{2} \\ k(k+1)=m^{2} \end{array}$$

The above equation holds only when $k=0$, hence we get
$$x_{0}=0, y_{0}=0$$

Therefore, the original equation has the unique solution $\left\{\begin{array}{l}x=0, \\ y=0 .\end{array}\right.$