There are $k^2 - (k-1)^2 - 1 = 2(k-1)$ non-square integers between $(k-1)^2$ and $k^2$.
So there are $2(1+2+3+\cdots+(k-1)) = k(k-1)$ non-square integers preceding the $k^2$-th square integer.
We want $k(k-1) = n \Rightarrow k = \frac{1+\sqrt{4n+1}}{2}$.
So the $n$-th non-square integer is $$n + \left[\frac{\sqrt{4n+1}+1}{2}\right]$$ where $[\cdot]$ is the floor function, i.e., $n + \left\lfloor\sqrt{n} + \frac{1}{2}\right\rfloor$.
It suffices to show that $$\langle\sqrt{n}\rangle = \begin{cases} \frac{\sqrt{4n+1}-1}{2} & \text{if it is an integer} \\ \left[\frac{\sqrt{4n+1}+1}{2}\right] & \text{if not} \end{cases}$$
(i) is an integer if and only if $n$ is of the form $m(m-1)$ for some $m \in \mathbb{Z}^+$, and equals $(m-1)$. To prove (i), we need $\langle\sqrt{m(m-1)}\rangle = m-1$; indeed $(m-1)-1/2 < \sqrt{m(m-1)} < (m-1)+1/2$.
For (ii), we observe $[\sqrt{n+1/4}+1/2] = [\sqrt{n}+1/2]$ for $n \in \mathbb{Z}$ and $n > 1$, which equals $\langle\sqrt{n}\rangle$.