Does $x+\sqrt{x}$ ever round to a perfect square, given $x\in \mathbb{N}$? I'll define rounding as $$R(x)=\begin{cases} \lfloor x \rfloor, & x-\lfloor x \rfloor <0.5 \\ \lceil x \rceil, & else\end{cases}$$

Does $x+\sqrt{x}$ ever round (to the nearest integer) to a perfect square, given $x\in \mathbb{N}$? 

For example, $7+\sqrt{7}=9.646...$ which rounds up, and $57+\sqrt{57}=64.549...$ which also round up. Also, $6+\sqrt{6}$ and $57+\sqrt{57}$ both round down. 
I think the positive integers $x$ such that $\lfloor x+\sqrt{x} \rfloor =k^2, k\in \mathbb{Z}$ are all of the form $n^2+n+1, n\in \mathbb{Z}^+$. The set of all $x$ begins as: $\{3, 7, 13, 21, 31, 43, 57, \dots \}$ and all those numbers are of the form $n^2+n+1$
I tried to find a pattern for whether the decimal part of $\sqrt{n^2+n+1}$ is less than $0.5$ or not, and I tried to modify $\sqrt{n^2+n+1}$ to $\sqrt{n^2+2n+1}=(n+1)^2$ but that didn't lead anywhere. 
Is there an algebraic proof/disproof of my above claim? 
Thanks. 

Assume $x\in\Bbb Z^+$.
If $x=m^2$ is a pefect square, then
$$m^2<x+\sqrt x=m^2+m<m^2+2m+1=(m+1)^2$$
an so $x+\sqrt x=R(x+\sqrt x)$ cannot be a perfect square.
Thus we need only consider the case that $x$ is not a perfect square, which makes $\lfloor x+\sqrt x\rfloor <\lceil x+\sqrt x\rceil$.
Let $n\in\Bbb Z^+$ be maximal with $n(n+1)<x$. Then $x=n^2+n+d$ with $1\le d\le (n+1)(n+2)-n(n+1)=2n+2$.
This makes
$$\lfloor x+\sqrt x\rfloor = n^2+n+d+n=(n+1)^2+d-1.$$
This is $\ge (n+1)^2$ and $\le n^2+4n+2<(n+2)^2 $. 
Hence $ k^2=\lfloor x+\sqrt x\rfloor$
implies $k=n+1$, $x=n^2+n+1$. But $$(n+\tfrac12)^2=n^2+n+\tfrac14<x$$ implies that $x+\sqrt x$ should round up, not down.
Similarly, $k^2=\lceil x+\sqrt x\rceil = \lfloor x+\sqrt x\rfloor+1$ implies  $k=n+2$, $d=(n+2)^2+1-(n+1)^2=2n+2$, $x=n^2+3n+2$. 
But $$ (n+\tfrac32)^2=n^2+3n+\tfrac 94>x$$
implies that $x+\sqrt x$ should be rounded down, not up.
We conclude that $R(x+\sqrt x)$ is never a perfect square for $x\in\Bbb Z^+$.
