jmc

Let $a=x,$ $b=2y,$ and $c=4z.$ Then $x=a,$ $y=\frac{1}{2}b,$ and $z=\frac{1}{4}c,$ so the given system becomes $$\begin{array}{rl}a+b+c& =12,\\ ab+ac+bc& =44,\\ abc& =48.\end{array}$$Then by Vieta's formulas, $a,$ $b,$ and $c$ are the roots of $$t^3-12t^2+44t-48=0.$$This factors as $(t-2)(t-4)(t-6)=0,$ so $a,$ $b,$ $c$ are 2, 4, 6, in some order.

There are $3!=6$ ways to assign 2, 4, 6 to $a,$ $b,$ and $c.$ These produce $\boxed{6}$ different solutions $(x,y,z),$ via the substitution $x=a,$ $y=\frac{1}{2}b,$ $z=\frac{1}{4}c.$