I posted this yesterday asking for help and was told it was unclear what I was asking. I continued to work on it and think I have completed the proof, so now I just am hoping I can get verification that the proof holds.
Proof: Prove by induction that if $w$ is an extended integer, that is $w = a+\sqrt{3}b$, and that $|N(w)| = w\cdot \bar{w} = a^2-3b^2 = 3^k$ for $k\geq 0$, then $N(w) = (-3)^k$.
We proceed with induction. We begin with the base case, $k=0$. We have $|N(w)| = 3^0 = 1$, which implies that $N(w) = \pm1$. We see that $N(w) = 1 = (-3)^0$ and we know, from past exercises, that there are no extended integers such that $N(w) = -1$ so the base case holds true for $k=0$.
Next, we assume the case for $n$ where $1\leq n \leq k$ to be true, that is, if $|N(w)| = 3^n$, then $N(w) = (-3)^n$ is true. Now we must prove that if $|N(w)| = 3^{n+1}$, then $N(w) = (-3)^{n+1}$ for $1\leq n+1 \leq k$. We have $N(w) = (-3)^{n+1} = a^2-3b^2$. Rearranging this equation, we are left with $a^2 = (-3)^{n+1}+3b^2 = 3((-3)^{n}-b^2)$. This implies, from previous exercises, that since $a^2$ is divisible by $3$, then $a$ is also divisible by $3$. We say that $a = 3t$, for some integer $t$, substituting $a$ into the $N(w)$ equation we are left with $N(w) = (3t)^2-3b^2 = 9t^2-3b^2 = -3(b^2-3t^2) = (-3)^{n+1}$. Factoring out a $-3$ from the right end of the above equation, we are left with $-3(b^2 - 3t^2) = -3(-3)^n$ and so, $b^2-3t^2 = (-3)^n$ which we already have assumed to be true, hence the statement is proven.