Here is a short proof of the more general identity

$$ \sum_{k=0}^{2m} (-1)^k \binom{2m}{k} \binom{x}{k}\binom{x}{2m-k} = (-1)^m \binom{2m}{m} \binom{x+m}{2m}. $$

Considered as polynomials in $x$, both sides have degree $2m$. If $x = m$ then $\binom{x}{k}\binom{x}{2m-k}$ is non-zero only when $k=m$, and so both sides equal $(-1)^m \binom{2m}{m}$. If $x \in \{0,1,\ldots,m-1\}$ then both sides are zero. If $x = -r$ where $r \in \{1,\ldots, m\}$ then, using that $\binom{-r}{k} = \binom{r+k-1}{r-1}$,
the left-hand side becomes

$$ \sum_{k=0}^{2m} (-1)^k \binom{2m}{k} f(k) $$

where $f(y) = \binom{r+y-1}{r-1} \binom{r+2m-y-1}{r-1}$. Since $f$ is a polynomial of degree $2(r-1) < 2m$ in $y$, its $2m$-th iterated difference is zero. So both sides are again zero. This shows that the two sides agree at $2m+1$ values of $x$, and so they must be equal as polynomials in $x$.

This identity is a specialization of (2) in the Gessel-Stanton paper linked to above.
It is (6.56) in Volume 4 of Gould's tables. To get Dixon's identity, take $x=2m$ and use that $\binom{2m}{2m-k} = \binom{2m}{k}$.