A tight result is $$\lceil \frac{n}{\sqrt{3}}\rceil \ge \frac{n^2}{\sqrt{3n^2-6+\frac{12}{n^2+2}}}$$

This means that the desired inequality holds for $n \ge 4.$ The cases $n=2,3$ can be checked separately.

As noted elsewhere, for $m=\lceil \frac{n}{\sqrt{3}}\rceil$ we have $n^2 \le 3m^2-2$ with equality exactly for the cases $n=1,5,19,71,265,989,\cdots$ A001834 with $\frac{n}{m} \lt \sqrt{3}$ a convergent to that square root. A bit of manipulation then gives the result above and the information that equality holds for exactly those values of $n.$

**Details**: Change the desired inequality to $$m^2 \ge \frac{n^4}{3n^2-k}$$ with $k$ to be determined. Since $n^2=3m^2-j$ with $j \ge 2,$ we have $m^2=\frac{n^2+j}{3}.$ This yields $$k \le \frac{3jn^2}{n^2+j} =3j-\frac{3j^2}{n^2+j}.$$ Since we do sometime have $j=2,$ The bound $k \le 6-\frac{12}{n^2+2}$ is sometime an equality. It is not hard to show that $\frac{3jn^2}{n^2+j}$ increases when $j \ge 2$ does.