The solutions are $(x, y) = (0, 0), (4, 5), (5, 4)$.
Let $a = x + y$ and $b = x - y$. The equation can be rewritten as
$$28(x + y) = 3((x + y)^2 + 3(x - y)^2),$$
which means $28a = 3(a^2 + 3b^2)$. Clearly, $3 \mid a$. Let $a = 3c$. The equation becomes
$$28c = 9c^2 + 3b^2.$$
We further find that $3 \mid c$. Let $c = 3d$. The equation becomes
$$28d = 27d^2 + b^2.$$
Now, as $d(28-27d) = b^2 \ge 0$, we must have $0 \le d \le \frac{28}{27}$. The only integer solutions are $d = 0, 1$.

When $d = 0$, $b = 0$. One finds that $a = 0$, and hence $x = y = 0$.

When $d = 1$, $b = \pm 1$. One finds that $a = 9$, and hence $(x, y) = (4, 5), (5, 4)$.

All these are solutions since the steps are reversible.