1. We need to find all natural number pairs $(x,y)$ satisfying $x^2-2^y=2023.$ So, $x^2=2023+2^y.$ We must identify all $(x,y)\in \mathbb N\times \mathbb N$ for which the right-hand side is a perfect square.

2. Rewrite the equation as a difference of squares: $x^2-2023=2^y.$ Since $2023$ is odd, let us factor $x^2-2023$ after expressing $2023$ suitably.

First factorize $2023$: $2023=7\cdot 17^2.$ But more useful here is to use parity and factorization directly: $x^2-2023=2^y.$ Since $2023$ is odd, $x^2$ must be odd $+$ even, hence odd. Therefore $x$ is odd. So $x-\sqrt{2023}$ is not useful over integers, but modular/factor methods are.

3. Since $x$ is odd, $x^2\equiv 1 \pmod 8$ for odd $x$. Also, $2023\equiv 7 \pmod 8.$ Thus $2^y=x^2-2023\equiv 1-7\equiv -6\equiv 2 \pmod 8.$ Now powers of $2$ modulo $8$ are:

• $2^1\equiv 2 \pmod 8$,
• $2^y\equiv 0 \pmod 8$ for all $y\ge 3$,
• $2^2\equiv 4 \pmod 8$. Hence the only possibility is $y=1.$

4. Substitute $y=1$ into the equation: $x^2-2=2023$ $x^2=2025$ $x=45$ (since $x\in\mathbb N$).

So the only pair is $(x,y)=(45,1).$

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

6. Comparison with stored answer: Our derived answer is $46$, which matches the stored correct answer.