@Roby5 @user11235813
This question can benefit from a CAS (Computer Algebra System) assistance.
As this kind of tools is still not widespread, I give here the Mathematica requests I have used (other CAS work along very similar lines)
g = GroebnerBasis[{$4z - 1 - (x + y)^2, 4x - 1 - (y + z)^2, 4y - 1 - (z + x)^2$}]
p = g[[1]]
Factor[p]
Answer:
$\{ 1241 - 4156\,z + 2044\,z^2 + 2016\,z^3 + 1136\,z^4 + 64\,z^5 + 64\,z^6,$
$ 17 - 918\,y + 879\,z - 648\,y\,z + 616\,z^2 - 216\,y\,z^2 + 272\,z^3 + 48\,z^4 + 16\,z^5,$
$-493 + 10368\,y + 2592\,y^2 - 8778\,z + 5184\,y\,z - 8360\,z^2 - 976\,z^3 - 528\,z^4 - 32\,z^5,$
$-3085 + 10368\,x + 10368\,y - 8778\,z - 10952\,z^2 - 976\,z^3 - 528\,z^4 -
32\,z^5\}$
$1241 - 4156\,z + 2044\,z^2 + 2016\,z^3 + 1136\,z^4 + 64\,z^5 + 64\,z^6$
${\left( -1 + 2\,z \right) }^2\,\left( 73 - 4\,z + 4\,z^2 \right) \,
\left( 17 + 12\,z + 4\,z^2 \right)$
Explanations : we have chosen the first polynomial in a list of 4, because it is the only one which is expressed with a single variable ($z$).
What conclusion can be drawn ?
That, necessarily, unknown $z$ should verify:
$$p(z):=
1241 - 4156\,z + 2044\,z^2 + 2016\,z^3 + 1136\,z^4 + 64\,z^5 + 64\,z^6=0$$
which can be given under the following factorized form:
$$p(z)=\left( -1 + 2\,z \right)^2\,\left( 73 - 4\,z + 4\,z^2 \right) \,
\left( 17 + 12\,z + 4\,z^2 \right)=0$$
where quadratic factors have no real roots.
Thus, necessarily $z=\dfrac{1}{2}$.
Due to the circularity of the given system, $x=y=\dfrac{1}{2}$ as well.
(immediate checking : $x=y=z=\dfrac{1}{2}$ is a solution...).
Remark: I don't give here any explanation about Groebner bases. IMHO, it is a kind of tool the existence of which should be known (the same is true for other algebraic tools like resultants, Sturm sequences, etc.) ; a deep understanding of the background theories is not necessary beyond a certain point in a first step. But most people having seen the power of these tools, their mathematical appetite being wetted, will sooner or later desire to understand "what is under the hood"...