Find a closed-form expression for $\sum_{k=0}^n\left[\binom{n}{k}\binom{m}{k}\right]$, where $m$ is fixed. Find a closed-form expression for $\sum_{k=0}^n\left[\binom{n}{k}\binom{m}{k}\right]$, where $m$ is fixed.

That sigma is messing me up!  How would I simplify this equation algebraically(I prefer solving it using algebra)?  If there isn't a way to solve it algebraically, could you provide a counting solution?  Thanks in advance!

Though you already have a nice algebraic answer, I'd like to offer a combinatorial answer, because I like them, and I hope you could appreciate them more by having one for this answer.
(For convention, I'll let $[n]=\{1,2...n\}$.) The left side counts the number of pairs $(A,B)$ where $A\subseteq [n], B\subseteq [m], |A|=|B|$. The right side counts size $m$ subsets of $[n+m]$. Now we partition $[n+m]$ into $[n]$ and $\{n+1, n+2...n+m\}$ (let's call it $[n+1,n+m]$), with the two sets having sizes $n$ and $m$ respectively.  Now, for each pair $(A,B)$ we can interpret $B$ as a subset of $[n+1,n+m]$ by identifying each $i\in B$ with $n+i\in [n+1,n+m]$ (this is clearly a bijection).
What we want now is a bijection between pairs $(A,B)$ with $A\subseteq [n]$, $B\subseteq[n+1,n+m]$, $|A|=|B|,$ and size $m$ subsets of $[n+m]$. Consider the set $[m]$: for a pair $(A,B)$, we look at $([m]- B)\cup A$. That is, we start with just the numbers greater than $n$, but then replace some of them with some numbers less than or equal to $n$, hence why we need $|A|=|B|$. Showing this to be a bijection is fairly straightforward, as each size $m$ subset of $[n+m]$ can be determined, and uniquely so, by which elements less than or equal to $n$ that is has, and which elements greater than $n$ that it's missing.
