Let $m, n\in \mathbb{N}$ and $x < 1$. I look for hints to derive an analytic formula for $$f_{m,n}(x) = \sum_{k \in \mathbb{N}} {n + k \choose k} {m + k \choose k} x^{k}. $$
3 Answers
This is Gauss' hypergeometric function $F(n+1,m+1,1;x)$. You can then apply the huge theory of hypergeometric functions to derive further expressions. For instance, Euler's transformation formula gives the alternative expression $$\frac 1{(1x)^{m+n+1}}\,F(m,n,1;x)=\frac 1{(1x)^{m+n+1}}\sum_{k=0}^{\min(m,n)}\binom mk\binom nk x^k$$ for the same series as a finite sum.

$\begingroup$ Fun fact: If one expands the RHS into powers of $x$, then one gets the equivalent binomial identity $\sum_{p+r=k}\binom{m+n+p}{p}\binom{m}{r}\binom{n}{r}=\binom{m+k}{k}\binom{n+k}{k}$. This identity is due to Surányi (1955), and it appears on Page 17 of Riordan: Combinatorial identities; make the substitution $(k,n,p,q)\mapsto(r,k,m,n)$ there. It provides an alternative proof of the expression for $f_{m,n}(x)$ in this post. $\endgroup$ Commented Nov 22, 2021 at 15:43

2$\begingroup$ @GHfromMO Unsurprisingly, the identity you attribute to Surányi is also a special case of a classical hypergeometric identity, namely, the PfaffSaalschütz summation. I believe Pfaff found it in the late 1700s. $\endgroup$ Commented Nov 22, 2021 at 15:58

4$\begingroup$ I'm reminded of the following quotation of Richard Askey (The work of George Andrews: a Madison perspective, 1999): "For years I had been trying to point out that the rather confused world of binomial coefficient summations is best understood in the language of hypergeometric series identities. Time and again I would find firstrate mathematicians who had never heard of this insight and who would waste considerable time proving some apparently new binomial coefficient summation which almost always turned out to be a special case of one of a handful of classical hypergeometric identities." $\endgroup$ Commented Nov 22, 2021 at 16:06

$\begingroup$ My main point was the existence of a more combinatorial proof, but I thank you for clarifying this story which lies outside my field of expertise. For fellow readers, the story can be found in more detail here: doi.org/10.1016/00973165(89)900885. See also doi.org/10.1016/00973165(85)900573 and the book Koepf: Hypergeometric summation  An algorithmic approach to summation and special function identities. $\endgroup$ Commented Nov 22, 2021 at 19:53

2$\begingroup$ I agree with this point. It is interesting to find different proofs, such as a combinatorial proof or the proofs shown in other answers. $\endgroup$ Commented Nov 23, 2021 at 5:24
With proofs abound, I would like to record yet another instance of application due to the WilfZeilberger techniques. The aim is to prove the identity shown above (from GH from MO, Hjalmar Rosengren): $$\sum_r\binom{m+n+kr}{kr}\binom{m}r\binom{n}r\binom{m+k}k^{1}\binom{n+k}k^{1}=1.$$ The mechanical process begins with letting (suppressing other variables) $$F(n,r):=\binom{m+n+kr}{kr}\binom{m}r\binom{n}r\binom{m+k}k^{1}\binom{n+k}k^{1}$$ and also that $$G(n,r):=\frac{F(n,r)\,(m+n+1+kr)r^2}{(m+n+1)(n+1+k)(nr+1)}.$$ Next step: verify $F(n+1,r)F(n,r)=G(n,r+1)G(n,r)$ which upon summing (both sides) over all integers $k$ results in cancellation on the RHS. Hence, $f(n+1)f(n)=0$ where $f(n)=\sum_kF(n,r)$. Since a direct computation shows $f(0)=1$, it follows that $f(n)=1$. The proof is now complete. $\,\,\square$
Hjalmar Rosengren gave a nice formula for $f_{m,n}(x)$ based on the theory of hypergeometric series. Here I provide a direct elementary proof of the same based on generating series.
Let us introduce the differential operator $$D_r:=\frac{1}{r!}\frac{\partial^r}{\partial x^r}.$$ Then $$\sum_{k=0}^\infty\binom{n+k}{k}x^k=D^n\left(\frac{1}{1x}\right)=\frac{1}{(1x)^{n+1}},$$ hence by the general Leibniz rule \begin{align*}\sum_{k=0}^\infty\binom{m+k}{k}\binom{n+k}{k}x^k &=D^m\left(\frac{x^m}{(1x)^{n+1}}\right)\\ &=\sum_{k=0}^m D^{mk}(x^m)D^k\left(\frac{1}{(1x)^{n+1}}\right)\\ &=\sum_{k=0}^m\binom{m}{k}\binom{n+k}{k}\frac{x^k}{(1x)^{k+n+1}}\\ &=\frac{1}{(1x)^{m+n+1}}\sum_{k=0}^m\binom{m}{k}\binom{n+k}{k}x^k(1x)^{mk}\\ &=\frac{1}{(1x)^{m+n+1}}\sum_{k=0}^m D^n(x^{n+k})\binom{m}{k}(1x)^{mk}. \end{align*} Here we can interpret $D^n(x^{n+k})$ as the coefficient of $t^n$ in the polynomial $(x+t)^{n+k}\in\mathbb{Z}[x][t]$, hence by the binomial theorem, the $k$sum on the righthand side equals $$[t^n]\sum_{k=0}^m (x+t)^{n+k}\binom{m}{k}(1x)^{mk}=[t^n](x+t)^n(1+t)^m.$$ Expanding $(x+t)^n$ and $(1+t)^m$ by the binomial theorem, and then collecting the coefficient of $t^n$ in the product of the resulting two sums, we get that $$\sum_{k=0}^\infty\binom{m+k}{k}\binom{n+k}{k}x^k=\frac{1}{(1x)^{m+n+1}}\sum_{k=0}^{\min(m,n)}\binom{m}{k}\binom{n}{k}x^k.$$

1$\begingroup$ Thanks this will be extremely useful for the kind of manipulations I need to do (in particular I need to take some partial sums). $\endgroup$– tomateCommented Nov 25, 2021 at 21:00