elementary number theory - Prove by induction that $2^{2n} – 1$ is divisible by $3$ whenever n is a positive integer.

I am confused as to how to solve this question.

For the Base case $n=1$, $(2^{2(1)} - 1)\,/\, 3 = 1$, base case holds

My induction hypothesis is:
Assume $2^{2k} -1$ is divisible by $3$ when $k$ is a positive integer

So, $2^{2k} -1 = 3m$

$2^{2k} = 3m+1$

after this, I'm not quite sure where to go. Can anyone provide any hints?