I don’t know whether you’d call them combinatorial, but here are two completely elementary arguments of a kind that I’ve presented in a sophomore-level discrete math course. Added: Neither, of course, is as nice as Didier’s, which I’d not seen when I posted this.
Let $b_n$ be the number of words of length $n$ with an odd number of $A$’s and an odd number of $B$’s, $c_n$ the number of words of length $n$ with an even number of $A$’s and an even number of $B$’s, and $d_n$ the number of words of length $n$ with an odd number of $A$’s and an even number of $B$’s. Then
$a_{n+1}=4a_n+b_n+c_n\;.\tag{1}$
Clearly $a_n=d_n$: interchanging $A$’s and $B$’s gives a bijection between the two types of word. $(1)$ therefore reduces to
$a_{n+1}=4a_n+b_n+c_n\;.\tag{2}$
But $6^n=a_n+b_n+c_n+d_n=b_n+c_n+2a_n$, so $b_n+c_n=6^n-2a_n$, and $(2)$ becomes $a_{n+1}=2a_n+6^n,a_0=0\;.\tag{3}$ $(3)$ can be solved by brute force:
$\begin{align*} a_n&=2a_{n-1}+6^{n-1}\\ &=2(2a_{n-2}+6^{n-2})+6^{n-1}\\ &=2^2a_{n-2}+2\cdot 6^{n-2}+6^{n-1}\\ &=2^2(2a_{n-3}+6^{n-3})+2\cdot 6^{n-2}+6^{n-1}\\ &=2^3a_{n-3}+2^2\cdot 6^{n-3}+2\cdot 6^{n-2}+6^{n-1}\\ &\;\vdots\\ &=2^na_0+\sum_{k=0}^{n-1}2^k6^{n-1-k}\\ &=6^{n-1}\sum_{k=0}^{n-1}\left(\frac13\right)^k\\ &=6^{n-1}\frac{1-(1/3)^n}{2/3}\\ &=\frac{2^{n-1}3^n-2^{n-1}}{2}\\ &=\frac{6^n-2^n}4\;. \end{align*}$
Alternatively, with perhaps just a little more cleverness $(1)$ can be expanded to
$\begin{align*} a_{n+1}&=4a_n+b_n+c_n\\ b_{n+1}&=4b_n+2a_n\\ c_{n+1}&=4c_n+2a_n\;, \end{align*}\tag{4}$
whence
$\begin{align*}a_{n+1}&=4a_n+4b_{n-1}+2a_{n-1}+4c_{n-1}+2a_{n-1}\\ &=4a_n+4a_{n-1}+4(b_{n-1}+c_{n-1})\\ &=4a_n+4a_{n-1}+4(a_n-4a_{n-1})\\ &=8a_n-12a_{n-2}\;. \end{align*}$
This straightforward homogeneous linear recurrence (with initial conditions $a_0=0,a_1=1$) immediately yields the desired result.