Let the set $$C = \{(x, y) \mid x^2 - 2^y = 2023, x, y \in \mathbb{N}\}$$. Then $$\sum_{(x,y) \in C}(x + y)$$ is equal to ______.

We need to find all natural number solutions $$(x, y)$$ of $$x^2 - 2^y = 2023$$.

This gives $$x^2 = 2023 + 2^y$$.

Check small values of y: for $$y = 1$$: $$x^2 = 2023 + 2 = 2025 = 45^2$$. ✓ So $$(45, 1)$$ is a solution.

For $$y = 2$$: $$x^2 = 2027$$, not a perfect square.

For $$y = 3$$: $$x^2 = 2031$$, not a perfect square.

For $$y = 4$$: $$x^2 = 2039$$, not a perfect square.

For $$y = 5$$: $$x^2 = 2055$$, not a perfect square.

For $$y = 6$$: $$x^2 = 2087$$, not a perfect square.

For $$y = 7$$: $$x^2 = 2151$$, not a perfect square.

For $$y = 8$$: $$x^2 = 2279$$, not a perfect square.

For $$y = 9$$: $$x^2 = 2535$$, not a perfect square.

For $$y = 10$$: $$x^2 = 3047$$, not a perfect square.

For $$y = 11$$: $$x^2 = 4071$$, not a perfect square.

For large y, use modular arithmetic: for $$y \geq 2$$: $$x^2 \equiv 2023 \pmod{4}$$. Since $$2023 = 505 \times 4 + 3$$, we get $$x^2 \equiv 3 \pmod{4}$$. But squares are $$\equiv 0$$ or $$1 \pmod{4}$$, which is a contradiction.

So no solutions exist for $$y \geq 2$$.

Compute the answer: the only solution is $$(x, y) = (45, 1)$$.

$$ \sum_{(x,y) \in C} (x + y) = 45 + 1 = 46 $$

The answer is $$\boxed{46}$$.