Question 1023501
The first few elements of A are 
{{{2^0+2 = 3}}}, {{{2^1+2 = 4}}}, {{{2^2+2 = 6}}}, {{{2^3+2 = 10}}}, {{{2^4+2 = 18}}}, {{{2^5+2 = 34}}},...
The first few elements of B are 
{{{2^1-2= 0}}}, {{{2^2-2 = 2}}}, {{{2^3-2 = 6}}}, {{{2^4-2 = 14}}}, {{{2^5-2 = 30}}},...

It is quite clear that 6 is a common element, so 6 is in A ∩ B.

Now we show that for k and {{{l>=4}}}, there are no other common terms.

Suppose there are, or suppose there are natural numbers a, {{{b>=4}}} such that 

{{{2^a + 2 = 2^b - 2}}}
==> {{{2^a + 4 = 2^b}}}, 
==> {{{2^(a-2)+1  = 2^(b-2)}}}, contradiction, because {{{2^(a-2)}}} and {{{2^(b-2)}}} would both be even since a, {{{b>=4}}}.

Therefore A ∩ B = {6}.