Solution:
Since $a x>0$ and $4^{x}-3^{x}>0$ it follows that $a>0$ and $x>0$.

For $a>0$, $x>0$ and $a x \neq 1$ the equation is equivalent to
$$3^{x}+4^{x}=7 \cdot 2^{x-1} \sqrt{4^{x}-3^{x}} \Longleftrightarrow 45\left(\frac{4}{3}\right)^{2x}-57\left(\frac{4}{3}\right)^{x}-4=0$$
Setting $y=\left(\frac{4}{3}\right)^{x}>0$ we obtain the equation $45 y^{2}-57 y-4=0$ with solutions $y_{1}=\frac{4}{3}$ and $y_{2}=-\frac{1}{15}$. Hence $x=1$. The condition $a x \neq 1$ now implies that $a \neq 1$.

The required values of $a$ are $a \in (0,+\infty) \setminus \{1\}$.