Proof by induction: $2^{2n}-1$ is a multiple of $3$ I'm learning proofs by induction and I'm a little confused on how they work exactly. This is what I have.
Theorem: $\forall n\in\mathbb N_0$,  $2^{2n}-1$ is a multiple of 3.
old proof with mistakes:

Base: $n=1$
$2^{2(1)}-1 = 4-1 = 3$
$3 = 3m, m\in\mathbb N$
$3$ is a multiple of $3$, so the theorem holds for the base case.
Step: $n\ge 2$
Induction hypothesis: $2^{2n}-1:=3m, m\in\mathbb N$
Induction conclusion: $2^{2(n+1)}-1=3m, m\in\mathbb N$
$2^{2(n+1)}-1 = 2^{2n+2}-1$
= $4*2^{2n}-1$
= $4*3m$ by the induction hypothesis
= $12m$
= $3(4m)$
$2^{2(n+1)} = 3m, m\in\mathbb N$

So $2^{2n}-1$ is a multiple of 3 $\forall\in\mathbb N$
Is the logic behind this correct?
Edit: corrections:

Step: n ≥ 2
Induction hypothesis: $2^{2n}-1=3m, m ∈ Z$
Induction conclusion: $2^{2(n+1)}-1=3m, m ∈ Z$
$2^{2(n+1)}-1 = 2^{2n+2}-1$
= $4*2^{2n}-1$
= $4*(2^{2n}-1)+3$
= $4*3m+3$ by the induction hypothesis
=$12m+3$
$2^{2(n+1)}-1=3(4m+1)$

$4m+1\in\mathbb N$, so $2^{2n}-1$ is a multiple of 3 $\forall n\in\mathbb N_0$.

You have applied the induction hypothesis wrongly. We have $4 (2^{2n}) -1=4 (2^{2n}-1)+3=4(3m)+3=3(4m+1)$
