On $x^3+y^3+z^3 = 1$ and a Pell equation

Question

Given,

$$(1-ac+bc)^3 + (a+c^2-ac^3)^3 + (ac^3-b-c^2)^3 = 1\tag{1}$$

where,

$$a,b,c,r = 12qrt,\;\; 3(q-r)(3q+r)t,\;\; 3s^2t^2,\;\; p-18qs^3t^3$$

then $(1)$ holds true if $p,q,s,t$ satisfies,

$$p^2-3(108s^6t^6-1)q^2=s\tag{2}$$

For $s=1$, this gives the Pell equation,

$$p^2-3(108t^6-1)q^2=1$$

which, starting with fundamental solution $p_1, q_1 = 216t^3-1,\; \pm12t^3$, gives an infinite family of polynomial parameterizations to $(1)$.

Question: Other than square $s$ and $s=3$, is there any other integer $s$ such that $(2)$ has a non-trivial solution in the integers?

Answer 1

Yes, there are other integers $s$. Just put $t=0$, so the equation becomes $p^2+3q^2=s$. Then pick any integers $p$ and $q$, and most likely $p^2+3q^2$ won't be either a square or $3$. For instance, if $p=2$ and $q=1$ then $s=7$.

If you require that $t$ be nonzero, then there are still solutions to (2) for other integers $s$ besides squares and $3$, for instance $s=12$ works with $t=1$, $q=2$, and $p=62208$.