See my response to Zach Teitler's comment. I was referring to the pencil of circles tangent to two given circles, maybe not exactly an Apollonius problem, but it is called Apollonian Circles.

You're absolutely right, sorry for not responding. I thought of another problem where the sought ellipse simply crosses the two other, but that's another job.

Nice reduction. I know well the problem of Apollonius, and I thought of a generalisation. I am currently working on a space flight project for an intensive foundation degree, where optimisation of fuel is the main objective. Thanks a lot, I will try to adapt this to the 3D case.

@ZachTeitler In fact I never mentioned that the two base ellipses were tangent, so the 3D problem is very general, and I don't think really obvious.

I found interesting rewriting $z=x+k$ and $c=b+i$ with $k,i\in\mathbb{N}^∗$. With your work, it becomes $x^{2b}(x^i+1)\equiv0\ [z]\ \mbox{and}\ k^b(k^i-1)\equiv0\ [x]$. Too bad I didn't think to congruences earlier.

I also analysed a bit the equation, and the solutions to the reduced single equation must verify $x<z$ and $c>b$ since $z^c=x^{b+c}+(x+z)^b>x^c$ and $z^c=x^{b+c}+(x+z)^b>z^b$.

OK, I edited my question, you may remove your vote.