NOTE: I do not want the eventual answerer to post a full solution; I only wish that they point me in the right direction as to how to do so.
Ok I am trying to prove this for $n>3$:
$$2\left(\sum_{r=1}^{2^{n-2}} \frac{(2^n)!}{(2r-1)!(2^n-(2r-1))!}\right) = 2 \left( \sum_{r=0}^{2^{n-2}-1} \frac{(2^n)!}{(2r)!(2^n-2r)!}\right)+\frac{(2^n)!}{((2^{n-1})!)^2}$$
For reference, it is also my theory that the sum of either side is $$2^{2^{n}-1}$$
The LHS has now been shown to be equal to the theorised sum. Many thanks to Richard for the detailed explanation (please see his answer for reference). I have since attempted to solve the RHS with his methods. Unfortunately, upon following his steps, midway through my efforts I was confronted by: $$\sum_{r=0}^{2^{n-2}-1} \binom{2^{n}-1}{2r-1}$$ - of which the notable problem is that for $r=0$ we get $$\binom{2^{n}-1}{-1}$$ which therefore leads to an obvious math error. Have I made an error here? If so can you point me in the right direction, or offer any relevant identities to show this? Any help is welcome. Thanks
EDIT: RESOLVED. I figured out how to work round it
First, observe that
$$\sum_{r=1}^{2^{n-2}}{\frac{2^n!}{(2r-1)!(2^n-(2r-1))!}} = \sum_{r=1}^{2^{n-2}}{\binom{2^n}{2r-1}}.$$
We then use the identity $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$:
$$\sum_{r=1}^{2^{n-2}}{\binom{2^n}{2r-1}} = \sum_{r=1}^{2^{n-2}}{\binom{2^n-1}{2r-2}} + \sum_{r=1}^{2^{n-2}}{\binom{2^n-1}{2r-1}} $$ $$= \sum_{r=0}^{2^{n-1}-1}{\binom{2^n-1}{r}} $$ $$=\frac{1}{2}\left( \sum_{r=0}^{2^{n-1}-1}{\binom{2^n-1}{r}} + \sum_{r=0}^{2^{n-1}-1}{\binom{2^n-1}{r}}\right) $$
Using the identity $\binom{n}{k} = \binom{n}{n-k}$, we can rewrite this as
$$=\frac{1}{2}\left( \sum_{r=0}^{2^{n-1}-1}{\binom{2^n-1}{r}} + \sum_{r=0}^{2^{n-1}-1}{\binom{2^n-1}{2^n-1-r}}\right) $$
$$= \frac{1}{2}\left( \sum_{r=0}^{2^{n-1}-1}{\binom{2^n-1}{r}} + \sum_{r=2^{n-1}}^{2^{n}-1}{\binom{2^n-1}{r}}\right) $$
$$= \frac{1}{2}\sum_{r=0}^{2^{n}-1}{\binom{2^n-1}{r}} = 2^{2^n-2}$$
where the final step follows from the identity $\sum_{k=0}^{n}{\binom{n}{k}}=2^n$. The RHS can be addressed similarly.