Squaring one of the zeroes of $f(x)=x^2-x-1$, if we write the zeros as $x=(1+q)$, where (needless to say), $q=x-1$, squaring said zero gives me $x^2=(2+q)$.
Stated differently, if $f(x)=0$, then the zeroes of $f$ are:
$$\left\lbrace\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2}\right\rbrace$$
Interestingly, numerically, the first zero is $1.618033989$, squaring it yields $2.618033989$.
Of course, it can be evaluated algebraically,, since: $x=\frac{1+\sqrt{5}}{2}$ can be written as: $$x=1+\frac{\sqrt{5}-1}{2}$$ $$x^2=\Biggl( 1+\frac{\sqrt{5}-1}{2} \Biggr)^2=1^2+2\cdot\frac{\sqrt{5}-1}{2}+\Biggl(\frac{\sqrt{5}-1}{2}\Biggr)^2$$ $$x^2=\frac{6}{4}+\frac{1}{2}\sqrt{5}=2+\frac{\sqrt{5}-1}{2}$$
Are there other numbers like this? Forgive my impudence.
If your question is about numbers keeping the same fractional part after squaring, it suffices to solve
$$x^2=x+n$$ where $n$ is an integer.
With $n=10$, $$(3.7015621187\cdots)^2=13.7015621187\cdots$$