Prove that $S_1:=\{z:0<\lvert z \rvert<R_1 \}$ and $S_2:=\{z:0<\lvert z \rvert<R_2 \}$ are conformally equivalent.
Proof: We need to find an analytic biholomorphic function $f:S_1\to S_2$. Consider $$f(z)=\frac{R_2}{R_1} e^z.$$ Then $|f(z)|=\frac{R_2}{R_1}e^x\le \frac{R_2}{R_1}R_1=R_2$. Also, for any $z_0\in S_2$, $\exists z_1\in S_1$ such that $\left| f(z_1) \right|= \left|z_0 \right|$, hence, $f$ is surjective. In addition, $f$ is injective, since for any $z_1, z_2\in S_1$ ($z_1\ne z_2$), $f(z_1)=\frac{R_2}{R_1}e^{z_1}\ne \frac{R_2}{R_1}e^{z_2}=f(z_2)$ Therefore, $S_1$ and $S_2$ are conformally equivalent.
I'd appreciate if you could give a feedback on my proof - whether or not it's correct. This is my first proof on conformalities.
As zhw commented, your map doesn't work. In general, $e^z$ is not injective, even on a punctured disc: $e^z = e^{z+2\pi i}$. Instead, you can use the map $f(z) = \frac{R_2}{R_1}z$ which has inverse $f^{-1}(z) = \frac{R_1}{R_2}z$.