Only if, the the equation ${{(b-z^2)}^{2}}-\frac{{{\left[ {{\left( a-z \right)}^{2}}+{{z}^{2}}-b \right]}^{2}}}{2}=c-{{z}^{4}}$ have infinite many solutions z! As the solution below shows, this is possible if, for example, a=0,b=c=2. (Remark: I denoted $x+y=S,xy=P$ so we have $S=a-z,{{S}^{2}}-2P=b-{{z}^{2}},{{({{S}^{2}}-2P)}^{2}}-2{{P}^{2}}=c-{{z}^{4}}$and I replaced S and P.)

It should be $(b-z^2)^2$ and $+z^2-b$ instead of $+z^2-1$ on your LHS. And actually your conclusion is quite wrong.

Not "quite" because you quoted my computations! Now I reformulated! Thank you for the honor!

It turns out one can actually describe all triples $(a,b,c)$ for which there are infinitely many solutions. (Credits to @Archimedes3.14159) A direct way to obtain these is as follows. Denote $p=xy+yz+zx$, $q=xyz$. Then $b = (x+y+z)^2 - 2(xy+yz+zx) = a^2 - 2p$ and $c = (x^2 + y^2 + z^2)^2 - 2(x^2y^2 + y^2z^2 + z^2x^2) = b^2 - 2(xy+yz+zx)^2 + 4xyz(x+y+z) = b^2 - 2p^2 + 4aq$. If $a\neq 0$, then given $(a,b,c)$ one expresses $p$ and $q$ uniquely and hence $x$, $y$, $z$ are uniquely determined as roots of the polynomial $X^3 - aX^2 + pX - q$ (either they exist as real numbers, with only one possibility for them as a set, or the system has no solutions). Thus for $a\neq 0$ we have (up to permutations) at most one possible $(x,y,z)$. On the other hand, if $a=0$, then $b= -2p = 2(x^2+xy+y^2)$ and $c = b^2 - 2p^2 = 2p^2 = 2(x^2+xy+y^2)^2$. These insist on $b,c > 0$ (as $x^2 + xy + y^2 \geq 0$ with equality only for $x=y=0$) and as long as $c = \frac{b^2}{2}$ we have infinitely many solutions, since $x^2 + xy + y^2 - \frac{b}{2} = 0$ has a real root for $x$ precisely when $D = 2b - 3y^2$, i.e. $|y| \leq \sqrt{\frac{2b}{3}}$ (which is an interval of infinitely many $y$).