I'm not sure where my work is wrong, I'm not obtaining an answer, even though I know there should be one.
In order to obtain the WZ proof certificate for the sum $$\sum_{k=0}^n \binom{n}{k} = 2^n$$ we have to first divide the right side: $$\sum_{k=0}^n \binom{n}{k}2^{-n} = 1$$ to set $F(n,k) = \binom{n}{k}2^{-n}$.
Now in order to have $$ F(n+1,k) - F(n,k) = G(n, k+1) - G(n,k)$$ we plug in our function: $$ \binom{n+1}{k}2^{-n-1} - \binom{n}{k}2^{-n} = G(n, k+1) - G(n,k)$$
This means we need to apply Gosper's algorithm to the left side $\binom{n+1}{k}2^{-n-1} - \binom{n}{k}2^{-n}$ in order to split it into its telescoping terms.
To apply Gosper's algorithm, we first define the ratio: $$r_k = \frac{t_{k+1}}{t_k} = \frac{ \binom{n+1}{k+1}2^{-n-1} - \binom{n}{k+1}2^{-n} }{ \binom{n+1}{k}2^{-n-1} - \binom{n}{k}2^{-n} } = \frac{ -(k-n-1)(2k-n+1) }{ (k+1)(2k-n-1) }$$
EDIT: found the mistake, fixed errors below!
Then we try to find polynomials $a(k)$, $b(k)$, $c(k)$ such that $r_k = \frac{a(k)}{b(k)}\frac{c(k+1)}{c(k)}$ and $\gcd(a(n), b(n+h)) = 1$ for all nonnegative integers $h$: $$\begin{align} a(k) &= -(k-n-1)\\b(k) &= k+1\\c(k) &= (2k-n-1) \end{align}$$
Substitute into the recurrence: $$a(k)x(k+1) - b(k-1)x(k) = c(k)\\ -(k-n-1)x(k+1) - kx(k) = 2k-n-1$$
This gives us $x(k) = -1$ and I am no longer stuck. Thanks for the help @Alexander Burstein.