Some algebra

Now consider a monic polynomial $g(t)$ having roots $x,y,z$ .We have that: $\begin{aligned} g(t)&=(t-x)(t-y)(t-z)\\ g(t)&=t^3-(x+y+z)t^2+(xy+xz+yz)t-xyz \end{aligned}$ We have that: $\begin{aligned} \color{#20A900}{x+y+z} &\color{#20A900}{=1}\\ (x+y+z)^2 &=1^2\\ x^2+y^2+z^2+2(xy+xz+yz)&=1\\ 35+2(xy+xz+yz)&=1\\ \implies \color{#D61F06}{xy+xz+yz}&\color{#D61F06}{=-17}\end{aligned}$ To find the value of $xyz$ , we make use of the identity $x^3+y^3+z^3-3xyz=\\(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$

Substituting the known values in,we get: $\begin{aligned} 97-3xyz&=(1)(35-(-17))\\ \implies 3xyz&=45\\ \implies \color{#3D99F6}{xyz}&=\color{#3D99F6}{15} \end{aligned}$ Substituting in the required values,we get: $g(t)=t^3-t^2+17t-15$ By the Rational Root Theorem ,we get that $t=5$ is a root of $g(t)$ .Factoring,we get $g(t)=(t-5)(t^2+4t+3)=(t-5)(t+1)(t+3)$ .Hence $(x,y,z)=\{5,-1,-3\}$ and $\text{max}(x,y,z)=\boxed{5}$

Let's suppose, without loss of generality, that $|x| \leq |y| \leq |z|$ . Let's also assume $-$ we shall justify this assumption later $-$ that $x$ , $y$ and $z$ are integers. In this case, since $1^2+3^2+5^2$ is the only way of writing $35$ as a sum of $3$ squares, it follows that $|x|=1$ , $|y|=3$ and $|z|=5$ . From the eight possible triples $(x, y, z)=(\pm 1, \pm 3, \pm 5)$ , only $(-1, -3, 5)$ further satisfies the other two equations, therefore max $(x, y, z)=\boxed{5}\,$ .

Let's now prove that there are no other solutions to the given system of equations, except for permutations of $(-1, -3, 5)$ . Its uniqueness follows from the fact that $z$ satisfies a cubic equation whose roots are $-1$ , $-3$ and $5$ . To derive such cubic equation, let's first combine the first two equations of the given system:

$(1-z)^2 = (x+y)^2 = x^2+y^2+2xy = 35-z^2+2xy \Rightarrow xy = z^2-z-17$ ;

this result, combined with all three equations, yields

$97-z^3 = x^3+y^3 = (x+y)(x^2-xy+y^2) = (1-z)[35-z^2-(z^2-z-17)] = (1-z)(52+z-2z^2) \Rightarrow z^3-z^2-17z-15 = 0.\, \square$