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

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 $x\in\Bbb Z^+$.

$$\sqrt{n^2+n+1}= \sqrt{\left(n+\frac12\right)^2+\frac 34} > n+\frac 12$$ And $$\sqrt{n^2+n+1} < \sqrt{n^2+2n+1} =n+1$$

$$\implies n+0.5 <\sqrt{n^2+n+1} <n+1$$

Thus $\rm{fractional part}{(n^2+n+1)}>0.5$