Positive integers $x, y, z$ satisfy $(x + yi)^2 - 46i = z$. What is $x + y + z$?
1. Start with the given equation:
   \[
   (x + yi)^2 - 46i = z
   \]

2. Expand the left-hand side:
   \[
   (x + yi)^2 = x^2 + 2xyi + (yi)^2 = x^2 + 2xyi - y^2
   \]
   Therefore, the equation becomes:
   \[
   x^2 + 2xyi - y^2 - 46i = z
   \]

3. Separate the real and imaginary parts:
   \[
   x^2 - y^2 + (2xy - 46)i = z
   \]

4. For the equation to hold, both the real and imaginary parts must be equal. Thus, we get two separate equations:
   \[
   2xy - 46 = 0
   \]
   \[
   x^2 - y^2 = z
   \]

5. Solve the first equation for \(xy\):
   \[
   2xy - 46 = 0 \implies 2xy = 46 \implies xy = 23
   \]

6. Find pairs \((x, y)\) that satisfy \(xy = 23\). The pairs are:
   \[
   (x, y) = (1, 23) \quad \text{or} \quad (x, y) = (23, 1)
   \]

7. Calculate \(z\) for each pair:
   - For \((x, y) = (1, 23)\):
     \[
     z = x^2 - y^2 = 1^2 - 23^2 = 1 - 529 = -528
     \]
   - For \((x, y) = (23, 1)\):
     \[
     z = x^2 - y^2 = 23^2 - 1^2 = 529 - 1 = 528
     \]

8. Since \(z\) must be a positive integer, we choose the pair \((x, y) = (23, 1)\), which gives \(z = 528\).

9. Calculate \(x + y + z\):
   \[
   x + y + z = 23 + 1 + 528 = 552
   \]

The final answer is \(\boxed{552}\).