Find all $z$ in the form $a + ib$ such that $\sin z = 2$. 
I need help on checking on my proof, especially on the part where i
  convert it using logarithm identities.

$\begin{aligned}
\sin z = 2 & \Rightarrow \dfrac{e^{iz} - e^{-iz}}{2i} = 2\\
 &\Rightarrow e^{iz} -e^{-iz} = 4i \\ 
 &\Rightarrow  e^{2iz} - 1  - 4ie^{iz} = 0\\
\end{aligned}$
Let $X = e^{iz}$, so that the above equation becomes $X^2 - 4iX -1 = 0$. Solve this quadratic equation to get $X = (2\pm\sqrt{3})i$.
It follows that $e^{iz} = (2\pm \sqrt{3})i = (2\pm \sqrt{3})e^{\pi i/2}$ as the argument of $(2+\sqrt{3})i$ is $90$ degrees.
Using the log identities, we can convert $2\pm\sqrt{3}$ to become $e^{\ln(2\pm \sqrt{3})}$
From here, we see that 
$\begin{aligned}
e^{iz} =e^{\ln(2\pm \sqrt{3})}e^{\pi i/2} &\Rightarrow iz =\ln(2\pm \sqrt{3})+\dfrac{\pi i}{2} + 2k\pi i\\
        &\Rightarrow z = -i\ln(2\pm\sqrt{3}) + \dfrac{\pi}{2}+2k\pi\\
\end{aligned}$

Yes, your proof is correct.  A few comments:
Note that because $(2 + \sqrt{3})(2 - \sqrt{3}) = 1$, we have $$-i \ln (2 \pm \sqrt{3}) = i \ln (2 \mp \sqrt{3}).$$
Another method of solution would be to use the angle addition identity to write $$ \sin (a+bi) = \sin a \cos bi + \cos a \sin bi = \sin a \cosh b + i \cos a \sinh b,$$
from which we conclude that $(a,b) \in \mathbb R^2$ must satisfy $$\sin a \cosh b = 2, \\ \cos a \sinh b = 0.$$  Since $\sinh b = 0$ if and only if $b = 0$, and $\cos a = 0$ if and only if $a \in \frac{\pi}{2} + \pi k$ for $k \in \mathbb Z$, we have two cases.  The first case is impossible:  if $b = 0$, then $\cosh b = 1$ and $\sin a = 2$ has no solutions for $a \in \mathbb R$.  So we must have instead $$\cosh b = 2 \csc \left(\frac{\pi}{2} + \pi k\right) = 2(-1)^k.$$  Since $\cosh b \ge 1$, it follows that $k$ must be even, and we get $b = \pm \cosh^{-1} 2$, and our solution set is $$a+bi = \frac{\pi}{2} + 2\pi m \pm i \cosh^{-1} 2, \quad m \in \mathbb Z.$$  It is worth observing that $\cosh^{-1} 2 = \ln(2 + \sqrt{3}).$
