Since the left side is an integer, the right side must also be an integer. The square root of an integer is either an integer or irrational (but never a non-integer fraction), so $\sqrt{x-y}$ must be an integer. The right side of the equation is divisible by 3, so the left side must also be. Thus, $3 \mid x^{2}+y^{2}$. But squares are always congruent to 0 or 1 modulo 3, so this can only happen if $x^{2} \equiv y^{2} \equiv 0 \bmod 3$. Therefore, $x$ and $y$ are both divisible by 3. Write $x=3 a$ and $y=3 b$ and substitute this in:

$$9 a^{2}+9 b^{2}+3^{3}=456 \sqrt{3 a-3 b}$$

Square roots of integers are generally either integers or irrational. But $\sqrt{3 a-3 b}$ cannot be irrational, because it is equal to $\frac{9 a^{2}+9 b^{2}+3^{3}}{456}$, so it must be an integer. Therefore, $3 a-3 b$ is a square, which is also divisible by 3, so it must be divisible by 9. We can now divide both sides of the equation by 9:

$$a^{2}+b^{2}+3=152 \sqrt{\frac{a-b}{3}}$$

Write $a-b=3 c^{2}$ and substitute $a=b+3 c^{2}$:

$$9 c^{4}+6 c^{2} b+2 b^{2}+3=152 c$$

Since all terms on the left side are positive, we must have $9 c^{4}+6 c^{2} b+2 b^{2}+3 \geq 152 c$. From $9 c^{4}+6 c^{2} b+2 b^{2}+3 \geq 152 c$ and $9 c^{4}+6 c^{2} b+2 b^{2}+3 \leq 152 c$, it follows that $c \leq 2$. Furthermore, $152 c$ is even, as is $6 c^{2} b+2 b^{2}$, so $9 c^{4}+3$ must also be even, which means $c$ must be odd. Therefore, the only possibility is $c=1$. Substituting this in, we get

$$9+6 b+2 b^{2}+3=152$$

or

$$b^{2}+3 b-70=0$$

This can also be written as $(b-7)(b+10)=0$. So $b=7$ or $b=-10$. In the first case, we get $a=b+3 c^{2}=10$ and thus $x=30$ and $y=21$. In the second case, we get $a=b+3 c^{2}=-7$ and thus $x=-21$ and $y=-30$. Checking shows that both pairs satisfy the equation. Therefore, the solutions are $(x, y)=(30,21)$ and $(x, y)=(-21,-30)$.

We are given the Diophantine equation:
$$x^2 + y^2 + 3^3 = 456\sqrt{x - y}$$

1. Introduce a new variable:
Let $n = \sqrt{x - y}$. Since $n$ must be an integer, $x - y$ must be a perfect square. Thus, we can write:
$$x = y + n^2$$

2. **Substitute $x = y + n^2$ into the original equation:**
$$(y + n^2)^2 + y^2 + 27 = 456n$$

3. Simplify the equation:
$$y^2 + 2y(n^2) + n^4 + y^2 + 27 = 456n$$
$$2y^2 + 2yn^2 + n^4 + 27 = 456n$$

4. Rearrange the equation:
$$2y^2 + 2yn^2 + n^4 - 456n + 27 = 0$$

5. Analyze divisibility by 3:
Since $3^3 = 27$, we have:
$$3 \mid y^2 + (y + n^2)^2$$
This implies $3 \mid y$ and $3 \mid y + n^2$. Therefore, $3 \mid n^2$, which means $3 \mid n$.

6. **Consider the parity of $n$:**
Since $n$ is divisible by 3 and must be odd, we have:
$$n \equiv 3 \pmod{6}$$

7. Express the equation in a different form:
$$(2y + n^2)^2 = -n^4 + 912n - 54$$
Let $P(n) = -n^4 + 912n - 54$. For $P(n)$ to be a perfect square, we need:
$$n(n^3 - 912) \leq -54$$
This implies:
$$0 < n < 10$$

8. **Check possible values of $n$:**
Given $n \equiv 3 \pmod{6}$, the possible values are $n = 3$ and $n = 9$.

9. **Evaluate $P(n)$ for $n = 3$ and $n = 9$:**
$$P(3) = -3^4 + 912 \cdot 3 - 54 = -81 + 2736 - 54 = 2601 = 51^2$$
$$P(9) = -9^4 + 912 \cdot 9 - 54 = -6561 + 8208 - 54 = 1593$$
Since $1593$ is not a perfect square, $n = 9$ is not a solution.

10. **Solve for $y$ when $n = 3$:**
$$2y + 3^2 = \pm 51$$
$$2y + 9 = 51 \quad \text{or} \quad 2y + 9 = -51$$
$$2y = 42 \quad \text{or} \quad 2y = -60$$
$$y = 21 \quad \text{or} \quad y = -30$$

11. **Find corresponding $x$ values:**
$$x = y + 3^2$$
$$x = 21 + 9 = 30 \quad \text{or} \quad x = -30 + 9 = -21$$

Thus, the integer solutions are:
$$(x, y) = (30, 21) \quad \text{and} \quad (x, y) = (-21, -30)$$

The final answer is $\boxed{(30, 21) \text{ and } (-21, -30)}$