Answer: D

Small Hint:

Compare exponents in $2^z=2\cdot4^x$ to express $z$ in terms of $x$

Big Hint:

For the intended positive solution, $z^x=y^{2x}$ gives $z=y^2$; substitute both relations into the sum

Solution:

For the intended positive integral solution, the first equation gives $z=y^2.$ The second equation is $$2^z=2^{2x+1},$$ so $z=2x+1$ and $x=\frac{y^2-1}{2}.$ Substituting into $x+y+z=16$ gives $$\frac{y^2-1}{2}+y+y^2=16.$$ The positive integral solution is $y=3,$ which gives $x=4$ and $z=9.$ Direct substitution verifies all three displayed equations.

Thus, the intended listed answer is D.