Find all pairs $(x, y)$ of integers that satisfy $x^2 + y^2 + 3^3 = 456\sqrt{x - y}$.
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 \(\boxed{(30, 21) \text{ and } (-21, -30)}\)