An induction proof should work well with this.
An observation to make is that powers of 2 alternate between congruencies depending if n is odd or even.
, if n is even
, if n is odd.
That is, divides 3 if n is even and divides 3 if n is odd.
Now, if x and n are integers, then 2^x and 3n are integers.
So look at the prime factorization of both sides. Clearly the prime factorization of the RHS contains a 3, but the prime factorization of the LHS contains only 2s. Thus the two sides cannot be equal for x and n integers.