Nicer proof that $2^{n+2}$ divides $3^m-1$ if and only if $2^{n}$ divides $m$

Given a prime $p$ and positive integer $k$, define $\nu_p(k)$ to be the exponent of $p$ in the prime factorization of $k$. We want to show that

$$\nu_2\left(3^n-1\right)=\begin{cases} 1&\mathrm{\ if\ }k\equiv 1\bmod 2\\ 2+\nu_2(n)&\mathrm{otherwise}.\end{cases}$$

We use strong induction to simplify calculations:

If $n$ is odd, then $3^n-1\equiv 2\bmod 4$, so the claim is true. Assume the claim is not true for all even $n$, and consider the smallest such $n=2m$. We have

$$3^{2m}-1=\left(3^m-1\right)\left(3^m+1\right).$$

If $m$ is even, then $\nu_2(3^m-1)=2+\nu_2(m)$ (by our strong inductive hypothesis) and $3^m+1\equiv 2\bmod 4$ so $\nu_2(3^m+1)=1$, which gives

$$\nu_2\left(3^{2m}-1\right)=3+\nu_2(m)$$

as desired. If $m$ is odd then $\nu_2(3^m-1)=1$ and $3^m+1\equiv 4\bmod 8$, so

$$\nu_2\left(3^{2m}-1\right)=1+2=3=2+\nu_2(2m),$$

as desired.

This isn't really different from your argument, it's just a more compact way of phrasing it. Similar results can be proven for any $p>2$; the most general form that looks nice is that if $p$ is an odd prime so that $p|a-b$ while $p\nmid a,b$, then

$$\nu_p\left(a^n-b^n\right)=\nu_p(a-b)+\nu_p(n).$$

(as user236182 mentioned, this is called Lifting the Exponent).

For $k\ge0$ and odd $p$, $$ 3^{p2^k}-1=\left(3^{2^k}-1\right)\overbrace{\left(3^{(p-1)2^k}+3^{(p-2)2^k}+\cdots+1\right)}^\text{$p$ terms} $$ For $k\ge2$, $$ 3^{2^k}-1=\left(3^{2^{k-1}}-1\right)\overbrace{\left(3^{2^{k-1}}+1\right)}^{2\pmod4} $$