Previously posted answers have amply dealt with the question, but I think it is worth making some methodological remarks.
The approach that is being taken to the question "When is $2^n >n^2$?" is essentially subtractive. There is a better approach, that deals well with a range of similar problems. For this particular problem, it will seem more complicated, because division feels more complicated than subtraction.
Let $f(n)=n^2/2^n$. We want to show that $f(n)<1$ when $n >4$. Note that
But $(1+1/n)^2/2<1$ for any $n \ge 3$. Thus from $n=3$ on, $f(n)$ is decreasing. Since $f(4)=1$, we conclude that $f(n)<1$ for all $n \ge 5$.
Similarly, let us show that from some identified point on, $3^n >n^4$. Let $f(n)=n^4/3^n$. A calculation similar to the one above shows that
The term that $f(n)$ is multiplied by is less than $1$ when $n \ge 4$, so from $n=4$ on, $f(n)$ is decreasing. A little experimentation shows that $f(7)>1$ and $f(8)<1$. So $f(n)<1$ from $n=8$ on.
Note for purists: The word "induction" has not been mentioned. However, like in most problems about positive integers, induction is being used. In the $2^n > n^2$ problem, from the facts that $f(n+1)<f(n)$ for $n \ge 3$ and that $f(4)=1$, we are quietly concluding that $f(n)<1$ for all $n>4$. In principle this step requires mathematical induction. In practice we regard this step as obvious, and do not use the "i" word.