Did you say infinite?

First, note that any square is either $1$ or $0 \pmod{4}$ . Now, suppose there exists a solution with $x$ being odd. Since the RHS is even, either $y$ or $z$ has to be odd and the other one even: $(2x‘-1)^{2}+(2y‘-1)^{2}+(2z‘)^{2}=2^{2}xyz‘$ and therefore $1+1+0\equiv 0 \pmod{4}$ which is a contradiction.

This means that neither $x$ , $y$ or $z$ can be odd, yielding $(2x‘)^{2}+(2y‘)^{2}+(2z‘)^{2}=2^{4}x‘y‘z‘$ and thus $x‘^{2}+y‘^{2}+z‘^{2}=2^{2}x‘y‘z‘.$

Since we can repeat the same argument for $x‘$ , $y‘$ and $z‘$ , they can only be $0$ , as any other finite value can not be devided by as many powers of $2$ .