Old question; the interesting application of Vieta Jumping is in several variables, for example Diophantine quartic equation in four variables, part deux and links there. The fundamental setup is
$$ x_1^2 + x_2^2 + \cdots x_n^2 = F(x_1, x_2, \ldots , x_n), $$
where $F$ is a symmetric polynomial for which every term is "squarefree" in each variable,all exponents are just $1.$ So, Markov Numbers are
$$ x^2 + y^2 + z^2 = 3 x y z. $$ Integral Apollonian circle packings are
$$ w^2 + x^2 + y^2 + z^2 = 2 w x + 2 w y + 2 w z + 2 x y + 2 x z + 2 y z. $$
The right hand side need not be homogeneous. A constant term is permitted, but that tends to throw things off because $(0,0,\ldots,0)$ is no longer a solution.
The earlier MSE question with this is
$$ w^2 + x^2 + y^2 + z^2 = wxyz -2 w x - 2 w y - 2 w z - 2 x y - 2 x z - 2 y z. $$
Notice how all variables (right hand side) appear without explicit visible exponents, so all exponents are actually $1.$
Anyway, what happens is that you get a forest of solutions, split up into rooted trees, possibly infinitely many. There are infinitely many in the Apollonian thing. Also, there, the roots include some negative numbers. By and large, it appears as though, when the right hand side $F$ has maximum degree higher than $2,$ we can probably split off the positive solutions into a finite number of trees.
EEDDIIIITT, May 24, 2014: This turned out to be interesting, Diophantine quartic equation in four variables asking for all solutions in positive integers to
$$ (w + x + y + z)^2 = wxyz. $$
While there are solutions with entries $0$ or negative, it turned out that we may split off the positive solutions, into exactly nine rooted trees. Go Figure.