$R$ is the ring $\mathbb{Z}[\sqrt{-k}]$ where $k\ge 5$ and $k\equiv2 \pmod{3}$.
I would like to prove that in $R$, if $3\mid(a+b\sqrt{-k})$, then $3\mid a$ and $3\mid b$ in $\mathbb{Z}$. I have proved the opposite direction, how can I prove this direction?
Also, I think $3$ is irreducible in $R$, is this correct and how can I prove that too?
The implication you're looking for is easy: if $a+b\sqrt{-k}=3(x+y\sqrt{-k})$, then $$ a=3x,\qquad b=3y $$ by equating the real and imaginary parts.
More interesting is asking whether $3$ is irreducible. Suppose $$ (a+b\sqrt{-k})(c+d\sqrt{-k})=3 $$ with both elements in the left hand sides noninvertible. Then, by taking conjugates, $$ (a^2+b^2k)(c^2+d^2k)=9 $$ so, since the elements are noninvertible, $a^2+b^2k=3$. Since $k\ge5$, this forces $b=0$.