Can you solve this by using induction? The inequality is true for $n = 1$, but is false until $n = 7$. After the induction step I got
$$2^n \geq n^2 + 2n + 1.$$
If you take the limit as $n$ approaches infinity, you see that $P(n+1)$ is also true.
My answer: The inequality holds true for $n \geq 7$.
The inequality amounts to $2^{n-1}>n^2$, from which you deduce $2^n> 2n^2$. To prove the inductive step, it suffices to prove that for $n\ge 7$, $\,\,2n^2\ge (n+1)^2$. This amounts to $p(n)=n^2-2n-1 \ge 0$ for $n \ge 7$. As $p$ has a positive and a negative root and $p(3)>0$, both roots are less than $3$, hence $p(n)>0$ for all $n \ge 3$.