Deletion of all bars in a word in $B^n$ produces a word in $A^n$. Now let $\alpha$ be a word in $A^n$ and let $a_i$ occur $k_i$ times in $\alpha$; the $k_i$ are positive even integers which add up to $n$. Since there are exactly $2^{k_i-1}$ distinct ways to bar $a_i$ an odd number of times in $\alpha$, the preimage of $\alpha$ under deletion of all bars has exactly $2^{k_1-1} \cdots 2^{k_m-1} = 2^{n-m}$ elements. The conclusion follows.

Alternative Solution:
The number $a_n$ is the coefficient of $x^n/n!$ in the formal expansion
$$A(x) = \left( \sum_{k \ge 1} \frac{x^{2k}}{(2k)!} \right)^m = \left( \frac{(e^x + e^{-x})}{2} - 1 \right)^m = 2^{-m} \left( e^{x/2} - e^{-x/2} \right)^{2m}.$$

Similarly, $b_n$ is the coefficient of $x^n/n!$ in the formal expansion
$$B(x) = \left( \sum_{k \ge 1} \frac{x^{2k-1}}{(2k-1)!} \right)^{2m} = 2^{-2m} (e^x - e^{-x})^{2m}.$$