4. Find all integer pairs $(x, y)$, such that
$$
3^{4} \times 2^{3}\left(x^{2}+y^{2}\right)=x^{3} y^{3} \text {. }
$$
4. If $xy=0$, then $x^2+y^2=0$. Hence $x=y=0$.

If $xy \neq 0$, then $x, y$ are either both positive integers or both negative integers. Since changing the signs of $x, y$ simultaneously does not change the original equation, without loss of generality, assume $x, y$ are both positive integers.

Let $x=3^m a, y=3^n b$, where $m, n \in \mathbf{N}$, and the positive integers $a, b$ are not divisible by 3.
Assume without loss of generality that $m \geqslant n$. Then the original equation becomes
$$
8\left(\left(3^{m-n} a\right)^{2}+b^{2}\right)=3^{3 m+n-4} a^{3} b^{3}.
$$

Since the remainder of a perfect square modulo 3 is 0 or 1, the left side of the above equation modulo 3 is not 0, which implies
$$
3 m+n-4=0.
$$

Since $m \geqslant n \geqslant 0$, we have $m=n=1$.
Thus, $8\left(a^{2}+b^{2}\right)=a^{3} b^{3}$.
By symmetry, assume without loss of generality that $a \geqslant b$. Then
$$
a^{3} b^{3}=8\left(a^{2}+b^{2}\right) \leqslant 16 a^{2}.
$$

Therefore, $a b^{3} \leqslant 16$.
If $b=2$, then $a=2$. Hence $(x, y)=(6,6)$.
If $b=1$, then $a \leqslant 16$, and $a^{3}=8\left(a^{2}+1\right)$, but no such positive integer exists.
In summary, the integer pairs that satisfy the condition are
$$
(x, y)=(-6,-6),(0,0),(6,6).
$$