A nice but somewhat challenging binomial identity 
When working on a problem I was faced with the following binomial identity valid for integers $m,n\geq 0$:
  \begin{align*}
\color{blue}{\sum_{l=0}^m(-4)^l\binom{m}{l}\binom{2l}{l}^{-1}
\sum_{k=0}^n\frac{(-4)^k}{2k+1}\binom{n}{k}\binom{2k}{k}^{-1}\binom{k+l}{l}
=\frac{1}{2n+1-2m}}\tag{1}
\end{align*}
I have troubles to prove it and so I'm kindly asking for support.

Maybe  the following  simpler one-dimensional identity could be useful for a proof. We have for non-negative integers $n$:
\begin{align*}
\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1}=\frac{4^{n}}{2n+1}\binom{2n}{n}^{-1}\tag{2}
\end{align*}
The  LHS of (2) can be transformed to
\begin{align*}
\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1}&=\sum_{k=0}^n(-1)^k\binom{n}{k}\int_{0}^1x^{2k}dx\\
&=\int_{0}^1\sum_{k=0}^n(-1)^k\binom{n}{k}x^{2k}\,dx\\
&=\int_{0}^1(1-x^2)^n\,dx
\end{align*}
Using a well-known integral representation of reciprocals of binomial coefficients the RHS of (2) can be written as
\begin{align*}
\frac{4^{n}}{2n+1}\binom{2n}{n}^{-1}&=4^n\int_{0}^1x^n(1-x)^n\,dx
\end{align*}
and the equality of both integrals can be shown easily. From (2) we can derive a simple one-dimensional variant of (1).
We consider binomial inverse pairs and with respect to (2) we obtain
\begin{align*}
&f_n=\sum_{k=0}^n(-1)^k\binom{n}{k}g_k \quad&\quad g_n=\sum_{k=0}^n(-1)^k\binom{n}{k}f_k\\
&f_n=\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1} \quad&\quad\frac{1}{2n+1}=\sum_{k=0}^n(-1)^k\binom{n}{k}f_k
\end{align*}

We conclude again with (2)
  \begin{align*}
\frac{1}{2n+1}&=\sum_{k=0}^n(-1)^k\binom{n}{k}f_k\\
&=\sum_{k=0}^n\frac{(-4)^{k}}{2k+1}\binom{n}{k}\binom{2k}{k}^{-1}\\
\end{align*}
  This identity looks somewhat like a one-dimensional version of (1). Maybe this information can be used to solve (1).


Firstly let us evaluate the inner sum on the left hand side. Using the beta function identity quoted above along with the identity $\left. \binom{k+l}{l} = d^l/dx^l x^{k+l}/l! \right|_{x=1}$ we have:
\begin{equation}
S^{(n)}_l:=\sum\limits_{k=0}^n \frac{(-4)^k}{2k+1} \binom{n}{k} [\binom{2k}{k}]^{-1} \binom{k+l}{l} = \left.\frac{1}{l!} \frac{d^l}{d x^l} x^l \int\limits_0^1 \left(1- 4 t (1-t) x\right)^n dt \right|_{x=1}
\end{equation}
Now if we take $m=0$ then $l=0$ and then:
\begin{equation}
rhs= 4^n \int\limits_0^1 \left[ (t-\frac{1}{2})^2 \right]^n dt= 4^n \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} u^{2 n} du = \frac{1}{2 n+1}
\end{equation}
as it should be.
Now let us take arbitrary $l \ge 0$ . Then by using the chain rule of differentiation and then by substituting $u := t-1/2$ we have:
\begin{equation}
S^{(n)}_l= \sum\limits_{p=0}^l \binom{l}{p} \binom{n}{p} \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} (4 u^2)^{n-p}(4 u^2-1)^p d u
\end{equation}
Therefore the left hand side of the identity to be proved reads:
\begin{eqnarray}
&&\sum\limits_{l=0}^m (-4)^l \binom{m}{l} [\binom{2 l}{l}]^{-1} S^{(n)}_l=\\
&&\sum\limits_{p=0}^m(-1)^{p+1} 2^{2p-1} \frac{\binom{m}{p} (m-p-3/2)!(p-1/2)!}{\sqrt{\pi} \binom{2 p}{p} (m-1/2)!} \binom{n}{p} \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} (4 u^2)^{n-p} (4 u^2-1)^p du=\\
&&\sum\limits_{p=0}^m (-1)^{p+1} 4^p \frac{\binom{m}{p} \binom{m}{1/2}}{\binom{2 p}{p} \binom{m}{p+3/2}} \cdot \frac{1}{(2p+1)(2p+3)} \binom{n}{p} \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} (4 u^2)^{n-p} (4 u^2-1)^p du=\\
&&-4^n \sum\limits_{p=0}^m \binom{m}{p} \frac{\binom{n}{p} \binom{m}{1/2}}{\binom{2 p}{p} \binom{m}{p+3/2}} \cdot \frac{1}{(2p+1)(2p+3)} \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} u^{2n-2p} (1-4 u^2)^p du=\\
&&-4^n \frac{1}{2} \frac{n!}{(m-1/2)!} \sum\limits_{p=0}^m  \binom{m}{p} \frac{(m-p-3/2)!}{(n-p)!} \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} (u^2)^{n-p} (1/4 - u^2)^p d u=\\
&& -\frac{1}{4} \frac{n! m!}{(n+1/2)! (m-1/2)!} \sum\limits_{p=0}^m \frac{(m-p-3/2)!(n-p-1/2)!}{(n-p)!(m-p)!}=\\
&& -\frac{1}{(2m-1)(2n+1)} F^{3,2}\left[\begin{array}{rrr} 1&-m&-n\\ \frac{3}{2}-m & \frac{1}{2}-n & \end{array};1\right] = \\
&& -\frac{1}{(2m-1)(2n+1)} \cdot \frac{(\frac{1}{2}-m)^{(n)} (\frac{3}{2})^{(n)}}{(\frac{3}{2}-m)^{(n)} (\frac{1}{2})^{(n)}} = \\
&&-\frac{1}{(2m-1)(2n+1)} \cdot  \frac{(1- 2m)(1+2 n)}{1-2 m+2 n} = \frac{1}{2n-2 m+1}
\end{eqnarray}
where in the first line we summed over $l$ and in the second in the third and in the forth  lines we simplified the result. Finally in the fifth line we evaluated the integral by substituting for $4 u^2$ and in the subsequent line we expressed the sum through hypergeometric functions. Finally, from Wolfram's site http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/03/02/04/,  we used the following identity:
\begin{equation}
F^{(3,2)}\left[ \begin{array}{rrr} a& b & -n \\ d & a+b-d-n+1 & \end{array};1 \right] = \frac{(d-a)^{(n)}(d-b)^{(n)}}{(d)^{(n)} (-a-b+d)^{(n)}}
\end{equation}
for $a=1$, $b=-m$ and $d=3/2-m$.
