<s>[INST] 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).

 [/INST] Let us complete the OP's work, started with
$$ \frac{1}{2k+1}\stackrel{\text{Binomial transform}}{\longleftrightarrow} \frac{4^k}{(2k+1)\binom{2k}{k}}\tag{$d=0$}$$
by computing first the binomial transform of $\frac{1}{2k+3}$. We have:
$$\begin{eqnarray*}\sum_{k=0}^{n}\frac{(-1)^k}{2k+3}\binom{n}{k}=\int_{0}^{1}x^2(1-x^2)^n=\frac{B\left(n+1,\tfrac{3}{2}\right)}{2}=\frac{1}{2n+3}\cdot\frac{B\left(n+1,\frac{1}{2}\right)}{2}\end{eqnarray*}$$
hence:
$$ \frac{1}{2k+3}\stackrel{\text{Binomial transform}}{\longleftrightarrow} \frac{4^k}{(2k+1)(2k+3)\binom{2k}{k}}\tag{$d=1$}$$
and in general:
$$ \frac{1}{2k+2d+1}\stackrel{\text{Binomial transform}}{\longleftrightarrow} \frac{4^k\binom{k+d}{d}\binom{2k}{k}^{-1}}{(2k+2d+1)\binom{2k+2d}{2d}}\tag{$d\geq 1$}$$
I need some time to check the above computations, but the last identity, together with creative telescoping, should be the key for proving OP's statement. Indeed, we have:
$$ \sum_{k=0}^{n}\frac{(-4)^k}{(2k+1)\binom{2k}{k}}\binom{n}{k}=\frac{1}{2n+1}\tag{$l=0$} $$
$$ \sum_{k=0}^{n}\frac{(-4)^k}{(2k+1)\binom{2k}{k}}\binom{n}{k}(k+1)=-\frac{1}{(2n+1)(2n-1)}\tag{$l=1$} $$
$$\begin{eqnarray*} \sum_{k=0}^{n}\frac{(-4)^k}{(2k+1)\binom{2k}{k}}\binom{n}{k}\binom{k+l}{l}&=&\frac{(-1)^l(2l-1)!!(2n-2l+1)!! }{(2n+1)!!}\\
&=&\frac{(-1)^l 4^{n-l} n! (2l)! (n-l)!}{(2n+1)!l! (2n-2l+1)!}\tag{$l\geq 1$} \end{eqnarray*}$$
hence the whole problem boils down to computing:
$$ \frac{4^n}{(2n+1)\binom{2n}{n}}\sum_{l=0}^{m}\frac{\binom{m}{l}}{(2n-2l+1)!\binom{n}{l}}$$
</s>