4. 165 Find the integer solution of the equation $\underbrace{\sqrt{x+\sqrt{x+\sqrt{x+\cdots+\sqrt{x}}}}}_{1964 \text { terms }}=y$.

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 translated part:
Find the integer solution of the equation $\underbrace{\sqrt{x+\sqrt{x+\sqrt{x+\cdots+\sqrt{x}}}}}_{1964 \text { terms }}=y$.

4. 165
[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 $\left\{\begin{array}{l}x=0, \\ y=0 .\end{array}\right.$