Notice that we can use the following property $\log_a \frac{b}{c}=\log_a b-\log_a c$ to get:
$$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)=\log_{yz}\frac{x^2+4}{4}-\log_{yz}\sqrt{yz}=\log_{yz}\frac{x^2+4}{4}-\frac{1}{2}$$
The equation can, thus, be written as:
$$\log_{yz}\frac{x^2+4}{4}+\log_{zx}\frac{y^2+4}{4}+\log_{xy}\frac{z^2+4}{4}=\frac{3}{2}$$
Notice that for any real number $a$, we have:
$$\frac{a^2+4}{4}\geq a \Leftrightarrow (a-2)^2\geq 0$$
with equality only if $a=2$. Therefore:
$$\log_{yz}\left(\frac{x^2+4}{4}\right)\ge \log_{yz} x=\frac{\ln x}{\ln y+\ln z}$$
Summing up, we arrive at:
$$\frac{3}{2}\geq \frac{\ln x}{\ln y+\ln z}+\frac{\ln y}{\ln x+\ln z}+\frac{\ln z}{\ln x+\ln y}$$
If we let $a=\ln x, b=\ln y, c=\ln z$, since $x,y,z>1$, we have $a,b,c>0$ and:
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq \frac{3}{2}$$
However, Nesbitt's inequality states that for any positive real numbers $a,b,c$, we must have:
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2}$$
with equality only if $a=b=c$. This implies immediately that the only solution of the equation is $x=y=z=2$.
Note: As far as I know, there is no standard way to solve this sort of multivariable logarithm equations.
Most of the time, you have to use inequalities and show that the equation is a particular equality case. In fact, it given that $x,y,z>1$ so that the logarithm will be positive ($\log_a b$ is positive if $a,b>1$ or $a,b<1$).
