First, let's consider the equation $$x^2 - 2^y = 2023$$ where $$x$$ and $$y$$ are natural numbers. Our goal is to find all the pairs $$(x, y)$$ that satisfy this equation and then sum the values of $$x+y$$ for each pair in set $$C$$.

Since $$2023$$ is an odd number, and $$x^2$$, the square of any natural number, is even when $$x$$ is even and odd when $$x$$ is odd, we can determine that for the left-hand side of the equation to be odd (thus equal to $$2023$$), $$x$$ must be odd since the right-hand side of the equation ($$2^y$$) is always even as it represents a power of two.

Also, $$2023$$ can be factored into prime factors to further analyze the possible solutions:

$$2023 = 7 \times 17 \times 17$$

Thus, allowing us to rewrite the equation as:

$$x^2 - 2^y = 7 \times 17^2$$

The next step is to check for potential values of $$x$$ that would fit the equation, keeping in mind that $$x$$ must be odd. We can try to express $$x^2$$ as $$7 \times 17^2$$ plus a power of $$2$$, recognizing that we are looking for the decomposition of the form:

$$x^2 = 7 \times 17^2 + 2^y$$

By examining the powers of $$2$$ and keeping in mind that they grow very quickly, we can reason that $$y$$ cannot be very large because $$x^2$$ must not exceed $$2023$$ by a large margin.

Let's start by trying the lowest values for $$y$$ since that would make $$2^y$$ small and $$x$$ has a better chance of being a natural number:

1. For $$y=1$$:

$$x^2 = 2023 + 2^1 = 2023 + 2 = 2025$$

Surprisingly, we find a perfect square since $$45^2 = 2025$$. Therefore, $$(x, y) = (45, 1)$$ is one solution.

1. For $$y=2$$ or higher:

$$2^y$$ becomes at least $$4$$ and increases exponentially, so $$x^2$$ must be at least $$2027$$ or higher in such cases. There's no natural number between $$45$$ and $$46$$, and $$46^2$$ far exceeds the target (2116), making it impossible for $$x^2$$ to be less than $$2116$$ for any larger $$y$$.

Hence, it appears there is only one possible solution: $$(x, y) = (45, 1)$$.

Therefore, the sum $$\sum_\limits{(x, y) \in C}(x+y)$$ for this set will consist of only this one pair:

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

So the answer is $$46$$.