If a number n is power of 2 AND n-1 is divisible by 3, THEN n must be power of 4 
If a number n is power of 2 AND n-1 is divisible by 3, THEN n must be power of 4

This holds for all the numbers I tried up to 32 bit maximum integer, but I wonder if it can actually be proven. The observation is that: If a number n is power of 2 AND n-1 is divisible by 3, THEN n must be power of 4.. For example: if n = 8 (not power of 4), n-1=7 is not divisible by 3, but if n = 64(power of 4) then n-1 = 63 is divisible by 3.
Mathematically speaking: if (2^k) - 1 = 3*m, for some positive integers m and k, then k must be an even number.
 A: If $k$ is odd then
$$2^k-1=(3-1)^k-1=(\text{sum of several multiples of }3)+(-1)^k-1=(\text{sum of several multiples of }3)-2$$
A: Well it is clear that every even power of two is a power of four. If you are familiar with modular arithmetic then if $n-1$ is divisible by 3 then this implies that $n\equiv1\pmod3$. Since $4\equiv1\pmod3$, then $4^x\equiv1\pmod3$ as well for some natural number $x$. If $x$ is even, then $2^x$ is a power of four and therefore $2^x\equiv1\pmod3$. If $x$ is an odd number then the remainder shall be two. So yes you are correct.
A: Assume $$(\exists n\in\mathbb N_0)(\forall m\in\mathbb N_0)\,\,\,2^n\neq4^m$$
Then $n$ must be odd.
$\implies$
$$(\exists k\in\mathbb N_0)\,\,\,2^n=2^{2k+1}=2(4^k)$$
Now $2(4^k)\equiv2(1^k)\equiv2\pmod3$
$\implies$
$$2^n\not\equiv1\pmod3$$
So we have proven
$$(\exists n\in\mathbb N_0)\,\,\,((\forall m\in\mathbb N_0)\,\,\,2^n\neq4^m\implies2^n\not\equiv1\pmod3)$$
The converse of this statement is
$$(\forall n\in\mathbb N_0)\,\,\,(2^n\equiv1\pmod3\implies (\exists m\in\mathbb N_0)\,\,\,2^n=4^m)$$
which is what we set out to prove.
