Can be rational? Can $z=\sqrt{1-4y^x}$ be a rational number, where $x>2$ is an integer and $y$ is a rational number and $y>0$?
I have tried with $x=2$ and it has rational solutions, for example $y=12/25$ and then $z=7/25$. But I haven't found any rational solution with $x>2$

Too long for a comment:
$\sqrt{1-4y^x}=z\quad=>\quad1-4y^x=z^2\iff1-4~\bigg(\dfrac mn\bigg)^x=\bigg(\dfrac ab\bigg)^2,\quad\begin{cases}\gcd\big(~a~,b\big)=1.\\\gcd\big(m,n\big)=1.\end{cases}$
$\iff\underbrace{n^x\overbrace{(b^2-a^2)}^{b~>~a}=\overbrace{4b^2m^x}^{>~0}}_{\gcd(a,b){\large=}1,\quad\gcd(m,n){\large=}1.}~=>~n^x$ can only be given by $4b^2$, and $b^2-a^2$ can only come from 
$4m^x$. Now, $4$ can be broken up into a product of two factors in only two ways: as $1\cdot4$, or as $2\cdot2$, 
giving us the three possible cases $~\begin{cases}~n^x=4b^2\\m^x=b^2-a^2\end{cases}\quad,~\begin{cases}~~~n^x=2b^2\\2m^x=b^2-a^2\end{cases}\quad,~\begin{cases}~~~n^x=b^2\\4m^x=b^2-a^2\end{cases}$
