N 2^N

0  1

1  2

2  4

3  8

4  7

5  5

6  1

7  2

8  4

9  8

10 7

11 5

and noting the cycle in the 2^N column it is clear that most numbers don't have a chance.  

In fact only if N is of the form 18a+5 will the digital roots of N and 2^5 be the same (5).

I could only check 5, 23, 41 on my calculator and only N=5 works.

It's possible, but unlikely there are other solutions, but the number of digits of 2^N grows as log(2) and the sum of digits grows as 4.5log(2)=1.35 it would be surprising to see another solution.