The original equation can be transformed into $\left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}=1$. Let $f(x)=\left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}$,
$\because f(x)$ is a monotonically decreasing function on $\mathbf{R}$, and $\left(\frac{3}{5}\right)^{2}+\left(\frac{4}{5}\right)^{2}=1$, i.e., $f(2)=1$,
$\therefore$ the original equation has one and only one solution $x=2$.