I believe the following identity holds for any non-negative integers $m,k$:
$$\binom{m+k}{k}^2 = \sum_{n=0}^m \binom{k}{m-n}^2\binom{2k+n}{n}.$$
It seems like there should be a slick double counting way of proving it, but I am not able to see it. I got this identity because I believe the following power series equality holds for $|x| < 1$,
$$(1-x)^{2k-1} \sum_{m=0}^\infty \binom{m+k-1}{k-1}^2 x^m = \sum_{j=0}^{k-1} \binom{k-1}{j}^2 x^j. $$
The binomial identity above comes from dividing by $(1-x)^{2k-1}$, applying the binomial theorem replacing $k$ with $k+1$ and comparing the coefficients of both sides.
But this power series equality doesn't seem any easier to prove than the binomial coefficient identity, since I don't really have a handle on the sums on either side.
Does anyone know a reference for an identity of this nature or have an idea for a double counting argument?