Prove $4^k - 1$ is divisible by $3$ for $k = 1, 2, 3, \dots$ For example:
$$\begin{align}
4^{1} - 1 \mod 3 &=
\\
4 -1 \mod 3 &= 
\\
3 \mod 3 &=
\\3*1 \mod 3 &=0
\\
\\
4^{2} - 1 \mod 3 &=
\\
16 -1 \mod 3 &= 
\\
15 \mod 3 &=
\\3*5 \mod 3 &= 0
\\
\\
4^{3} - 1 \mod 3 &=
\\
64 -1 \mod 3 &= 
\\
21 \mod 3 &=
\\3*7 \mod 3 &=
 0\end{align}
$$
Define $x = \frac{4^k - 1}{3}$. So far I have:
$$k_1 \to 1 \Longrightarrow x_1 \to 1
\\
k_2 \to 2 \Longrightarrow x_2 \to 5
\\
k_3 \to 3 \Longrightarrow x_3 \to 21
\\
k_4 \to 4 \Longrightarrow x_4 \to 85$$
But then it's evident that
$$4^{k_n} = x_{n+1} - x_n$$
I don't know if this helps, these are ideas floating in my head. 

Hint:
$$4 \equiv 1 \mod 3 \Rightarrow 4^n \equiv 1^n \mod 3 \Rightarrow 4^n \equiv 1 \mod 3$$
