Closed form for $\sum_{k=0}^{l}\binom{k}{n}\binom{k}{m}$ Does there exist any closed form for the following sum?

$$\sum_{k=0}^{l}\binom{k}{n}\binom{k}{m}$$

Where 
$l \in \mathbb N$ and $m,n \in \mathbb Z$

My try:
$$ \sum_{k=\max\left(m,n\right)}^{l}\binom{k}{n}\binom{k}{m}=\sum_{k=0}^{l}\binom{k}{k-n}\binom{k}{k-m}$$$$=\left(-1\right)^{\left(-n-m\right)}\sum_{k=0}^{l}\binom{-n-1}{k-n}\binom{-m-1}{k-m}$$$$=\left(-1\right)^{\left(-n-m\right)}\sum_{k=0}^{l}\binom{-n-1}{-1-k}\binom{-m-1}{k-m}$$$$=\left(-1\right)^{\left(-n-m\right)}\binom{-n-m-2}{-m-1}$$
$$=\left(-1\right)^{\left(-n-m\right)}\binom{-n-m-2}{-n-1}=\left(-1\right)^{\left(-m-1\right)}\binom{m}{-n-1}$$$$=\left(-1\right)^{\left(-m-1\right)}\binom{m}{m+n+1}=\left(-1\right)^{n}\binom{n}{m+n+1}$$
I'm not sure whether it's right, so can someone verify the solution, and if it's not right then please provide a closed form (of course if that's exist).

We present a proof of the identity by @Diger, which should be  considered
a starting point for additional simplification. We seek to show that
$$\sum_{k=0}^l {k\choose m} {k\choose n} =
\sum_{k=0}^n (-1)^k {l+1\choose m+k+1} {l-k\choose n-k}.$$
The RHS is
$$[z^n] \sum_{k=0}^n (-1)^k {l+1\choose m+k+1} 
z^k (1+z)^{l-k}.$$
The coefficient extractor enforces the range:
$$[z^n] \sum_{k\ge 0} (-1)^k {l+1\choose l-m-k} 
z^k (1+z)^{l-k}
\\ = [z^n] (1+z)^l [w^{l-m}] (1+w)^{l+1}
\sum_{k\ge 0} (-1)^k w^k
z^k (1+z)^{-k}
\\ = [z^n] (1+z)^l [w^{l-m}] (1+w)^{l+1}
\frac{1}{1+wz/(1+z)}
\\ = [z^n] (1+z)^{l+1} [w^{l-m}] (1+w)^{l+1}
\frac{1}{1+z+wz}
\\ = [z^n] (1+z)^{l+1} [w^{l-m}] (1+w)^{l+1}
\frac{1}{1+z(1+w)}
\\ = [z^n] (1+z)^{l+1} [w^{l-m}] 
\sum_{k\ge 0} (-1)^k z^k (1+w)^{k+l+1}
\\ = [z^n] (1+z)^{l+1}
\sum_{k\ge 0} (-1)^k z^k {k+l+1\choose l-m}.$$
This is
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=0}^n (-1)^k {l+1\choose n-k} {k+l+1\choose l-m}.}$$
The LHS is 
$$\sum_{k\ge 0} [[0\le k\le l]] [z^m] (1+z)^k [w^n] (1+w)^k
\\ = [z^m] [w^n]  \sum_{k\ge 0} (1+z)^k  (1+w)^k
[v^l] \frac{v^k}{1-v}
\\ = [z^m] [w^n] [v^l] \frac{1}{1-v} 
\sum_{k\ge 0} (1+z)^k  (1+w)^k v^k
\\ = [z^m] [w^n] [v^l] \frac{1}{1-v} 
\frac{1}{1-(1+z)(1+w)v}
\\ = [z^m] [w^n] 
[v^l] \frac{1}{v-1} 
\frac{1/(1+z)/(1+w)}{v-1/(1+z)/(1+w)}.$$
The inner term is
$$\mathrm{Res}_{v=0} \frac{1}{v^{l+1}}
\frac{1}{v-1} \frac{1/(1+z)/(1+w)}{v-1/(1+z)/(1+w)}.$$
Residues sum to zero and the residue at infinity in $v$ is zero.
The contribution from minus the residue at $v=1/(1+z)/(1+w)$ is
$$- [z^m] (1+z)^{l+1} [w^n] (1+w)^{l+1} 
\frac{1/(1+z)/(1+w)}{1/(1+z)/(1+w)-1}
\\ = - [z^m] (1+z)^{l+1} [w^n] (1+w)^{l+1} 
\frac{1/(1+z)}{1/(1+z)-(1+w)}
\\ = [z^m] (1+z)^{l+1} [w^n] (1+w)^{l+1} 
\frac{1/(1+z)}{w+z/(1+z)}
\\ = [z^m] (1+z)^{l+1} [w^n] (1+w)^{l+1} 
\frac{1/z}{w(1+z)/z+1}.$$
Now with $l,m,n$ positive integers we must have $l\ge n,m$ or else 
there is no contribution to $k^\underline{m} k^\underline{n}.$
This means we continue with
$$[z^m] (1+z)^{l+1} 
\sum_{k=0}^n {l+1\choose k} 
\frac{1}{z} (-1)^{n-k} \frac{(1+z)^{n-k}}{z^{n-k}}
\\ = \sum_{k=0}^n (-1)^{n-k} {l+1\choose k} 
{l+1+n-k\choose m+1+n-k}.$$
This is
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=0}^n (-1)^{n-k} {l+1\choose k} 
{l+1+n-k\choose l-m}.}$$
We have the same closed form for LHS and RHS, thus proving the claim.
 For a full proof we also need to show that the contribution from 
$v=1$ is zero. We get
$$[z^m] [w^n] \frac{1/(1+z)/(1+w)}{1-1/(1+z)/(1+w)}
= [z^m] [w^n]  \frac{1}{(1+z)(1+w)-1}
\\ = [z^m] [w^n]  \frac{1}{z+w+zw}
= [z^{m+1}] [w^n] \frac{1}{1+w(1+z)/z}
\\ = [z^{m+1}] (-1)^n \frac{(1+z)^n}{z^n}
= (-1)^n {n\choose n+m+1} = 0.$$
