Shall we try teamwork?  Please feel free to edit this post if you have simplifications.

The original sum may be re-expressed as
$$ \frac{1}{2^{2m+1}} \sum_{k=0}^m (-1)^k \binom{m}{k} \binom{2(k+m)}{k+m} \frac{1}{2^{2k}} \sum_{j=0}^{k+m-1} \frac{2^{k+m-j}}{(k+m-j) \binom{2(k+m-j)}{k+m-j}}. $$
If we're trying to prove this is 0, we may drop the fraction out front.  Also, change variables from $j$ to $\ell=k+m-j$:
$$ \sum_{k=0}^m \left( -\frac14 \right)^k \binom{m}{k} \binom{2(k+m)}{k+m} \sum_{\ell=1}^{k+m} \frac{2^\ell}{\ell \binom{2\ell}{\ell}}. $$
At this point, my idea was to change the order of summation based on
$$ \sum_{k=0}^m \sum_{\ell=1}^{k+m} \Diamond = \sum_{\ell=1}^m \sum_{k=0}^m \Diamond + \sum_{\ell=m+1}^{2m} \sum_{k=\ell-m}^m \Diamond, $$
but I can't get quite it to work out.  The first sum simplifies, but the second sum I can't do much with.

Any ideas?