combinatorics - Combinatorial proof of $sum limits_{i = 0} ^{m} 2^{n-i} {n choose i}{m choose i} = sumlimits_{i=0}^m {n + m - i choose m} {n choose i}$

I've been wondering for a while how to solve (prove) a combinatorial identity, using just combinatorial interpretation:

$$\sum \limits_{i = 0} ^{m} 2^{n-i} {n \choose i}{m \choose i} = \sum\limits_{i=0}^n {n + m - i \choose m} {n \choose i}$$

($m \leq n$ )

The left hand side is pretty much about choosing any number of elements from the set $M = \{a_1, \dots, a_m\}$ and then choosing at least the same amount from $N = \{b_1, \dots, b_n\}$, but I can't see how the right hand side satisfies that.