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!

Notice $\displaystyle\;\binom{m}{k} = 0$ whenever $k > m$. WOLOG, we only need to study the case $m \ge n$.
Given any polynomial $f(x) = \sum\limits_{k=0}^{\deg f} a_k x^k$, we will use $[x^k] f(x)$ to denote $a_k$, the coefficient of $x^k$ in $f(x)$.
By Binomial theorem, we have
$$\binom{n}{k} = [x^k] (1+x)^n\quad\text{ and }\quad
\binom{m}{k} = \binom{m}{m-k} = [x^{m-k}] (1+x)^m$$
This implies
$$\sum_{k=0}^n \binom{n}{k}\binom{m}{k}
= \sum_{k=0}^n \left([x^k](1+x)^n\right)\left([x^{m-k}](1+x)^m\right)
= [x^m] (1+x)^{n+m} = \binom{n+m}{m}
$$
