A system with finitely many solutions Let $a$, $b$ be (complex) numbers (satisfying some natural conditions, see below). Show that for every $(u,v,w)$ the system
$$ a x + b y + z = u \\ a x^2 + b y^2 + z^2 = v \\ a x^3 + b y^3 + z^3 = w$$
has finitely many solutions $(x,y,z)$.
The conditions on $a$, $b$, $c$ :  $a$, $b$, $a+1$, $b+1$, $a+b$, $a+b+1 \ne 0$ ( otherwise it is easy to produce infinitely many solutions by having some of the $x$, $y$, $z$ equal).
Note: We want to show that the above polynomial map in $(x,y,z)$ has finite fibers. It is clear how to generalize this statement. That is in fact equivalent to : the map is proper.  This perhaps can be checked in this particular case and in other cases using Gröbner bases. I am interested in an approach that might work in general.
$\bf{Added:}$ With WA I found an algebraic dependence of $x$ on $u$, $v$, $w$ (as it should). The coefficient of $x^6$ is  $a^2 (1 + a) (a + b) (1 + a + b)^2$ Dividing by it, we get that $x$ is integral over the algebra generated by $u$, $v$, $w$. The same works perhaps with $y$, $z$. The calculations are fairly large.
$\bf{Added:}$ Sketch of a proof: Stratify  $\mathbb{C}^n$  using equality of components. Assume that some fibre is infinite. It follows that its dimension is positive. But then the fibre restricted to some stratum has positive dimension. However, using the Jacobian, we see that restricted to a stratum the map $\Phi$ is locally injective. Contradiction
 A: There is a computer-based proof done with Mathematica 13.1.
The command
Solve[{a x + b y + z == u , a x^2 + b y^2 + z^2 == v, 
a x^3 + b y^3 + z^3 == w}, {x, y, z},  Assumptions -> 
a != 0 && b != 0 && a + 1 != 0 && b + 1 != 0 && a + b != 0 && a + b + 1 != 0]

produces the solution where $x,y,z$ are expressed as roots of polynomials with coefficients from the field of rational functions  $\mathbb Q (a,b,u,v,w)$. The whole output takes 79.6 MB in memory. Here is the screen of the shortened output. BTW, the execution of the command is not long.
Addition.The code
a = 1; b = -1; Reduce[{a x + b y + z == u, a x^2 + b y^2 + z^2 == v, 
  a x^3 + b y^3 + z^3 == w}, {x, y, z}]


(u^2 - v !=  0 && (x == ( 4 u^3 - 4 w -  Sqrt[(-4 u^3 + 4 w)^2 -  4 (6 u^2 - 6 v) (u^4 + 3 v^2 - 4 u w)])/(12 (u^2 - v)) ||  x == (4 u^3 - 4 w +  Sqrt[(-4 u^3 + 4 w)^2 -  4 (6 u^2 - 6 v) (u^4 + 3 v^2 - 4 u w)])/(12 (u^2 - v))) &&  y == (-u^3 + 3 u v - 2 w)/(3 (u^2 - v)) &&  z == u - x + y) || (w == 0 && v == 0 && u == 0 && x == 0 &&  z == y) || ((v == w^(2/3) || v == -(-1)^(1/3) w^(2/3) ||  v == (-1)^(2/3) w^(2/3)) && v != 0 && u == w/v && x == u &&  z == y) || (w == 0 && v == 0 && u == 0 && -x != 0 && y == x &&  z == -x + y) || ((v == w^(2/3) || v == -(-1)^(1/3) w^(2/3) ||  v == (-1)^(2/3) w^(2/3)) && v != 0 && u == w/v && u - x != 0 &&  y == x && z == u - x + y)

, as we see, results in infinite number of the solutions.
