Put n=2k because 2 | n

We have:2^2k-1=4^k-1

Because 4\(\equiv1\left(mod3\right)\)\(\Rightarrow4^k\equiv1^k=1\left(mod3\right)\Rightarrow4^k-1⋮3\)

\(\Rightarrow2^n-1\) is a multiple of 3 then 2 | n

2ⁿ = 2.2.2....2 with n factor 2

we see: 

2 split 3 residual 2

2.2 = 4 split 3 residual 1

2.2.2 = 8 split 3 residual 2

.........................................

So if n is even number, 2ⁿ will split 3 residual 1

=> 2ⁿ - 1 is a multiple of 3

So 2ⁿ - 1 is a multiple of 3 then 2 | n