For what $N$ does $2^N$ overflow?
You don't explain how $2^N$ is calculated, so I'm assuming you repeatedly multiply by 2 (or equivalently, you repeatedly add the number to itself, or shift it left). Here's how this process looks on a 4-bit signed register. You start with $0001$, the answer for $N=0$. Doubling it, we reach $0010$, the answer for $N=1$. Doubling again, we reach $0100$, the answer for $N=2$. Doubling again, overflow occurs (why?).

If the register was unsigned, we could double one more time, obtaining $1000$, the answer for $N=3$. Then we would overflow when doubling again (why?).

I believe you now have the tools to solve your own exercise. I won't verify your answer, so no point asking about it below.