$$\sum_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k$$ This is the exercise 3.3.6 of the book Invitation to Discrete Mathematics.
The answer in book is
Let M be an m-element set and N be an n-element set. Both sides count the number of ordered pairs $$(X,Y)$$ with $$X\subseteq M, Y\subseteq N\cup X,\ and\ |Y|=m$$ For the right-hand side, we first pick $$Y\cap N,$$ then $$Y\cap M$$ and finally X.
I'm confusing in the "finally X".
Change the order of summation i.e., $k'=m-k$ to $$\sum _{k=1}^{m}\binom{n}{m-k}\binom{m}{k}2^{m-k},$$ so $|Y\cap N|=m-k$ and $|Y\cap M|=k$, then notice that you have $m-k$ left from $M$ just choose some set, say $Z$, where doesn't matter the length but is such that $X=Y\cup Z$. The only requirement is that $Z\subseteq M\setminus Y$ and there are $2^{|M\setminus Y|}=2^{m-k}$ options.