Solution. We introduce the substitution $\log _{3} x=y$, from which we get $\sqrt{x}=\sqrt{3^{y}}$. Substituting into the original equation, we obtain

$$\begin{aligned} & \log _{2}\left(1+\sqrt{3^{y}}\right)=y \\ & 1+\sqrt{3^{y}}=2^{y} \\ & \left(\frac{1}{2}\right)^{y}+\left(\frac{\sqrt{3}}{2}\right)^{y}=1 \end{aligned}$$

An obvious solution to the last equation is $y=2$. If $y \geq 2$, then

$$\left(\frac{1}{2}\right)^{y}\left(\frac{1}{2}\right)^{2}=\frac{1}{4} \text { and }\left(\frac{\sqrt{3}}{2}\right)^{y}>\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{3}{4}$$

from which

$$\left(\frac{1}{2}\right)^{y}+\left(\frac{\sqrt{3}}{2}\right)^{y}>\frac{1}{4}+\frac{3}{4}=1$$

Thus, $y=2$ is the only solution to the equation $\left(\frac{1}{2}\right)^{y}+\left(\frac{\sqrt{3}}{2}\right)^{y}=1$, from which, by returning to the substitution, we get that $x=9$ is the only solution to the original equation.