$\lim_{k \to \infty} \frac{x_k}{k^2}$

Let $l=\lim_{k\to\infty}\frac{x_k}{k^2}$. Then by the Stolz-Cesàro theorem, \begin{eqnarray} l=\lim_{k\to\infty}\frac{x_k}{k^2}=\lim_{k\to\infty}\frac{x_{k+1}-x_k}{(k+1)^2-k^2}=\lim_{k\to\infty}\frac{\sqrt x_k}{2k+1}=\lim_{k\to\infty}\sqrt{\frac{x_k}{k^2}}\frac{k}{2k+1}=\frac12\sqrt{l} \end{eqnarray} and hence $l=0$ or $l=\frac14$. Note that $x_1>\frac1{16}$. Assume that $x_k>\frac{k^2}{16}$. Then $$ x_{x+1}=x_k+\sqrt x_k>\frac{k^2}{16}+\frac{k}{4}=\frac{k^2+4k}{16}>\frac{(k+1)^2}{16}. $$ Thus $x_k>\frac{k^2}{16}$ for $k\ge 1$ and hence $\frac{x_k}{k^2}>\frac1{16}>0$. So you can rule out $l=0$ and hence $$ l=\lim_{k\to\infty}\frac{x_k}{k^2}=\frac14. $$

Write $y_k=2\sqrt{x_k}$. Then the given equation becomes $y_{k+1}^2+1=(y_k+1)^2$.

Now prove by induction that $\sqrt{k^2-1}-1 \leq y_k \leq k+1$. Thus, $y_k/k \to 1$ and $x_k/k^2 \to 1/4$.