First note that if $xy = 0$, then $x^2 + y^2 = 0$ and $x = y = 0$ is a solution. Also note that if $xy < 0$, then $x^2 + y^2 < 0$, impossible; thus $x, y$ are both positive or both negative. Changing simultaneously the sign of $x$ and $y$ does not change the equation, hence we may assume without loss of generality that $x, y$ are both positive.
Let $m, n$ ($m \ge n$) be non-negative integers, and let $a, b$ be coprimes integers, not divisible by $3$. Consider $x = 3^m a$ and $y = 3^n b$. Then the given equation is
$$8((3^{m-n}a)^2 + b^2) = 3^{3m+n-4}a^3b^3.$$
Because any perfect square has remainder $0$ or $1$ when divided by $3$, the left hand side member is not divisible by $3$, thus looking at the right hand side we get $3m + n - 4 = 0$, and because $m \ge n \ge 0$ it follows $m = n = 1$. The equation takes the form
$$8(a^2 + b^2) = a^3 b^3.$$
By symmetry we may assume $a \ge b$. Then
$$16a^2 \ge a^3 b^3 \Leftrightarrow 16 \ge ab^3$$
Hence we have either (1) $b=2$ which implies $a=2$, or (2) $b=1$, but in this case the only possible values of $a$ are $1, 2, 4, 8$ and none of them satisfies the equation $a^3-8a^2-8=0$. It is easy to see that $(a, b) = (2, 2)$ and $(x, y) = (6, 6)$. So, the only solutions are
$$(x, y) = (-6, -6), \quad (x, y) = (0, 0), \quad (x, y) = (6, 6).$$