- First part of the task is just to show that $(4^n-1)$ actually is divisible by 3 for n=1,2,3,4. No problem.
Second step: is to show that $(4^n -1) = (2^n-1)(2^n+1)$ No problem, just algebra.
Third step is to explain that $(2^n-1)$,$2^n$,$(2^n+1)$ is three consecutive numbers. And that only one is divisible by three.
$2^n$ has two as a factor and is not divisible by 3. It can't be even. That leaves $(2^n-1)$ and $(2^n+1)$. The one with 3 as a factor seems to be random dependent on n.
Fourth step, tie it all together.
Now, I'm not sure if I'm suppose to dig deeper into which of them is actually divisible by 3. Or has 3 as a factor. But knowing that one of them at the time (dependent on n) has indeed 3 as a factor implies that, in respect to the second step, 3 is a factor of $(4^n -1)$.
Is this all there is to it? Keep in mind that this is a really beginners proof task, not very formal. It's slightly funny going back to high school math after being scared to death by logic and descrete math at university-level. I just expect everything thing to be super complex.