The answer is $k=3$. First, let $k=3$ and set $S_n = x^n+y^n+z^n$. Then $S_2^2 = S_3S_1\ge S_2^2$ by Cauchy-Schwarz, so $x=y=z$. Then, $3x=3x^2$ yielding $(x,y,z)=(1,1,1)$ as the unique solution. Now let $k=2$. We show $x+y+z=x^2+y^2+z^2$ has infinitely many solutions. Choose $z=1-y$ and consider this as a quadratic in $x$: \[ x^2 -x + (y^2+z^2-y-z)=0 \Rightarrow x^2-x +(1-y)^2 +y^2-1=0. \]The discriminant of the resulting quadratic is $1-8y^2-8y$, which is strictly positive for $y$ small enough, yielding infinitely many solutions.