In $\mathbb{Z}[\sqrt{-5}]$
I have worked out $a^2=\langle 2\rangle$ then $a^3=\langle 4,2+2\sqrt{-5}\rangle$ and $a^4=\langle 4\rangle$
Is it correct to say that $a^n=\langle 2^{n-1}\rangle$ for even numbers but I'm not sure about odd numbers.
I'm looking for $a^n$ in the simplest form.
If $a=\langle r,s\rangle$, then $$ a^2=\langle r^2,rs,s^2\rangle $$ so in your case $a^2=\langle 4,2+2\sqrt{-5},-4+2\sqrt{-5}\rangle$. Then $2\sqrt{-5}\in a^2$ and also $2\in a^2$. This (up to some small details) proves that indeed $a^2=\langle 2\rangle$.
Now $a^3=\langle 4,2+2\sqrt{-5}\rangle$(you shouldn't just cube the generators of $a$) and $a^4=\langle 4\rangle$. Can you then use induction after having made a suitable conjecture?