Suppose $ (a,c)=1$. We'll look for solutions with $ y=y_0^a$ and $ x=y_0^b$, so that $ x^a=y_0^{ab}=y^b$. Then we're solving $ 2x^a=z^c$. Now let $ x=2^uX$ and $ z=2^mZ$, so that $ X,Z$ are odd. Then $ 2^{1+au}X^b=2^{mc}Z^c$ and we must have $ 1+au=mc$ and $ X^a=Z^c$. Because $ (a,c)=1$, $ 1+au=mc$ has infinitely many solutions $ (u,m)$ in positive integers: just take $ m$ to be $ ka+c'$, where $ cc'\equiv1($ mod $ a)$, for any $ k>0$. This $ c'$ exists, as $ (a,c)=1$. Also $ X^a=Z^c$ is obviously solvable: take $ X=r^c$ and $ Y=r^a$, for some $ r>0$ odd. In this case we obtain $ x=2^ur^c$. Because $ x$ must be of the form $ y_0^b$, we need $ r$ to be a power of a positive integer with exponent multiple of $ b$ and $ u$ to be multiple of $ b$. For $ r$ there's no problem, take $ r=s^b$, for some $ s$ odd. For $ u$ it's a bit more complicate. Remember $ u$ has the form $ ka+c'$. Now the condition $ (a,b)$ helps. It ensures the existence of a positive integer $ 1\le a'<b$ such that $ aa'\equiv1($ mod $ b)$. We may take $ k=lb-a'c'$ for any $ l$ great enough to ensure that $ k>0$. Then $ ka+c'$ will be a multiple of $ b$, as needed. We can even neglect $ X,Z$ (that is, set them equal to $ 1$) and obtain solutions with $ x,y,z$ all powers of $ 2$.

freemind wrote: Remember $ u$ has the form $ ka+c'$. I thought that was $m$. I don't think you can get a solution with all powers of two if $(a,c) = 1$ but $(b,c) \neq 1$.

We can suppose $a$ and $c$ are relatively prime without loss of generality. Let $s$,$t$,$k$ be positive integers such that $k = s^{c}-t^{b}$. Since $a$ and $bc$ are relatively prime, there exist infinity many positive integers $w$ such that $w \equiv -1 \pmod{a},w \equiv 0 \pmod{bc}$ by Chinese Remainder Theorem. We take that $x = k^{\frac{w+1}{a}}$, $y = k^{\frac{w}{b}} \cdot t$, $z = k^{\frac{w}{c}} \cdot s$, then they satisfy the equation $x^{a}+y^{b} = z^{c}(=k^w \cdot s^{c})$. So,we have done.