Symmetric sum and also geometric sum

Since this is a geometric progression, we can write an equation based on the common ratio:

$\dfrac{x^2+y^2+z^2}{x+y+z} = \dfrac{x^3+y^3+z^3}{x^2+y^2+z^2}\\ (x^2+y^2+z^2)^2 = (x+y+z)(x^3+y^3+z^3)\\ {\color{#D61F06}x^4+y^4+z^4}{\color{#20A900}\,+\,2x^2y^2}{\color{#EC7300}\,+\,2x^2z^2}{\color{#3D99F6}\,+\,2y^2z^2 }= {\color{#D61F06}x^4+y^4+z^4}{\color{#20A900}\,+\,x^3y+xy^3}{\color{#EC7300}\,+\,x^3z+xz^3}{\color{#3D99F6}\,+\,y^3z+yz^3}\\ {\color{#20A900}x^3y-2x^2y^2+xy^3}{\color{#EC7300}\,+\,x^3z-2x^2z^2+xz^3}{\color{#3D99F6}\,+\,y^3z-2y^2z^2+yz^3} = 0\\ {\color{#20A900}xy(x^2-2xy+y^2)}{\color{#EC7300}\,+\,xz(x^2-2xz+z^2)}{\color{#3D99F6}\,+\,yz(y^2-2yz+z^2)} = 0\\ {\color{#20A900}xy(x-y)^2}{\color{#EC7300}\,+\,xz(x-z)^2}{\color{#3D99F6}\,+\,yz(y-z)^2} = 0$

Note that since $x \neq y \neq z$ , we know that $\color{#20A900}x-y \neq 0$ , $\color{#EC7300}x-z \neq 0$ and $\color{#3D99F6}y-z \neq 0$

None of the numbers $x,y$ and $z$ are $0$ . Suppose $x= 0$ . Then, $\color{#3D99F6}yz(y-z)^2 = 0$ , which implies that $y=x=0$ or $z=x=0$ , since $\color{#3D99F6}y-z \neq 0$ . This contradicts the fact that $x \neq y \neq z$ . The same argument holds if $y=0$ or $z=0$

Then, suppose that $x,y,z > 0$ . We know that ${\color{#20A900}xy > 0, (x-y)^2 > 0}\,,\, {\color{#EC7300}xz > 0, (x-z)^2 > 0}\,,\, {\color{#3D99F6}yz > 0, (y-z)^2 > 0}$ . This implies that the expression ${\color{#20A900}xy(x-y)^2}{\color{#EC7300}\,+\,xz(x-z)^2}{\color{#3D99F6}\,+\,yz(y-z)^2} >0$ .

Hence, $\boxed{0}$ distinct positive triplets $(x,y,z)$ satisfy the given requirements