Recall that given,
$$x^3+y^3+z^3 = (z+1)^3$$
we do the substitution $z = 3 n^2x + (3 n^2 + 1)(y - 1)$ like Adam Bailey in this post to get the ellipse,
$$x^2 - x y + y^2 - (27 n^4 - 1) x - (27n^4 + 18 n^2 + 2 ) y + (27n^4 + 9 n^2 + 1 ) = 0$$
after removing a trivial factor. I found two triples of polynomial solutions to this. The first triple $(x,y) = (a_j,a_k)$,
\begin{align} (a_1,\, a_2) &= (3n + 3n^2 + 9n^3,\, 1 + 6n^2 - 9n^3)\\ (a_3,\, a_4) &= (-3n + 3n^2 - 18n^3 + 27n^4,\, 1 - 3n + 15n^2 - 9n^3 + 54n^4)\\ (a_5,\, a_6) &= (12n^2 + 9n^3 + 54n^4,\, 1 + 3n + 15n^2 + 18n^3 + 27n^4)\end{align}
These three lattice points are the vertices of the first triangle. And the second triple $(x,y) = (b_j,b_k)$ ,
\begin{align} (b_1,\, b_2) &= (-9n^3,\, 1 - 3n + 9n^2 - 18n^3 + 27n^4)\\ (b_3,\, b_4) &= (-3n + 9n^2 - 9n^3 + 54n^4,\, 1 + 18n^2 + 9n^3 + 54n^4)\\ (b_5,\, b_5) &= (3n + 9n^2 + 18n^3 + 27n^4,\, 1 + 3n + 9n^2 + 9n^3)\qquad\qquad\end{align}
are the vertices of the second triangle. Naturally, these satisfy,
$$(a_j)^3+(a_{j+1})^3+(u_j)^3 = (u_j+1)^3$$ $$(b_j)^3+(b_{j+1})^3+(v_j)^3 = (v_j+1)^3$$
for integer polynomials $u_j,\,v_j.$ I found these empirically, but the natural question to ask is: can we derive one triangle from the other? Note the curious relationships,
\begin{align} \alpha &= a_1 + a_3 + a_5 = b_1 + b_3 + b_5 = 3(27n^4+6n^2)\\ \beta &= a_2 + a_4 + a_6 = b_2 + b_4 + b_6 = 3(27n^4+12n^2+1)\end{align}
\begin{align} \frac23(\alpha+\beta)\, &= a_1 + a_2 + b_3 + b_4\\ &= a_3 + a_4 + b_5 + b_6\\ &= a_5 + a_6 + b_1 + b_2 \,=\, 2(54n^4+18n^2+1)\end{align}
In a previous post, I asked if $(\alpha,\beta)$ had a geometric interpretation and Jan-Magnus Økland answered that $\big(\dfrac{\alpha}3,\dfrac{\beta}3\big)$ in fact is the center of the ellipse. And this post, it is stated that "if the center is known, then 3 points are enough to uniquely define an ellipse". A triangle, in other words.
To illustrate, let $n=\pm2$ and we get the same ellipse,
$$x^2 - x y + y^2 - 431 x - 506 y + 469 = 0$$
graphed below by the Desmos calculator with relevant lattice points,
The 3 blue points determine the first triangle, while the 3 green points determine the second triangle. By reflecting these $6$ points along the semi-major and semi-minor axes, then one will get $6\times4 = 24$ points. (However, the Alpertron calculator says this has a total of $48$.)
Question: Is there a general way to derive the second triangle from the first triangle using some basic principles of conic sections?
P.S. The reason I ask is I found three families of ellipses for $x^3+y^3+z^3=(z+1)^3$, each with a pair of triangles with polynomial vertices. It can't be coincidence they come in pairs.
(I made a tentative answer last night which helped in coalescing the complete answer. This is the revised version. There are three families of ellipses.)
Ellipse 1: Factoring $\,x^3+y^3+z^3 = (z+1)^3,\,$ where $z = 3 n^2x + (1 + 3 n^2)(y - 1).$
After removing a trivial factor, we get the ellipse in the post. An initial lattice point is:
$$(x,\,y) = (a_1,\;a_2) = (3n + 3n^2 + 9n^3,\; 1 + 6n^2 - 9n^3)$$
Starting with the single polynomial $a_1$, and the method described by Adam Bailey in this answer, one can find six $a_k$ (given in the post above) yielding the three lattice points,
\begin{align} (x_1,\,y_1) &= (a_1,\;a_2)\\ (x_2,\,y_2) &= (a_3,\;a_4)\\ (x_3,\,y_3) &= (a_5,\;a_6)\end{align}
which determine the vertices of the first triangle and,
\begin{align} \alpha &= a_1 + a_3 + a_5 = 3(27n^4+6n^2)\\ \beta &= a_2 + a_4 + a_6 = 3(27n^4+12n^2+1)\end{align}
which determine the center of the ellipse. But how to find the second triangle from basic principles? My first clue was Jan-Magnus Økland's moving the center of the ellipse to the point of origin $(0,0)$ using the transformation, $$x = p+q+\dfrac{\alpha}3,\quad y = p-q+\dfrac{\beta}3$$
to go from the long, $$x^2 - x y + y^2 - (27 n^4 - 1) x - (27n^4 + 18 n^2 + 2 ) y + (27n^4 + 9 n^2 + 1 ) = 0$$
to the much simpler, $$\color{blue}{p^2+3q^2} = 9n^2(3n^2 + 1)(27n^4 + 9n^2 + 1)$$
The form $p^2+3q^2$ is well-known, was my second clue, and is also discussed in Bailey's post here. If we can express $M$ by the form, then it can be done so in three ways,
$$M = p^2+3q^2 = \left(\frac{p+3q}2\right)^2+3\left(\frac{p-q}2\right)^2 = \left(\frac{p-3q}2\right)^2+3\left(\frac{p+q}2\right)^2$$
Since we know an initial point $(x_1, y_1)$, it is trivial to solve for $(p,q)$ from,
$$x_1 = p+q+\dfrac{\alpha}3,\quad y_1 = p-q+\dfrac{\beta}3$$
to get,
$$p = \frac32(n - 3n^2 - 18n^4),\quad q= \frac32(n + n^2 + 6n^3)$$
and we can get one vertex $(b_1, b_2)$ of the second triangle,
\begin{align} b_1 &= \left(\frac{p-3q}2\right)+\left(\frac{p+q}2\right)+\frac{\alpha}3 = -9n^3\\ b_2 &= \left(\frac{p-3q}2\right)-\left(\frac{p+q}2\right)+\frac{\beta}3 = 1 - 3n + 9n^2 - 18n^3 + 27n^4\end{align}
Using for the same process for the $a_i$, all three vertices of the second triangle can be found,
\begin{align} (x_4,\,y_4) &= (b_1,\;b_2)\\ (x_5,\,y_5) &= (b_3,\;b_4)\\ (x_6,\,y_6) &= (b_5,\;b_6)\end{align}
hence the second triangle was derived from the first, and we have answered our question. Incidentally,
$$a_1^3+a_2^3+z^3 = (z+1)^3$$
where $z=6n^2 + 27n^4,$ so this $z$ is immune to sign changes of $n$.
Ellipse 2: Factoring $\,x^3+y^3+z^3 = (z+1)^3,\,$ where $z = 36n^4x + (1 + 36n^4)(y - 1).$
Initial lattice points:
\begin{align}(x,\,y) = (c_1,\;c_2) =\; &(6n^2 - 36n^3 + 36n^4 + 648n^6 - 1296n^7,\\ & 1 + 12n^2 + 72n^4 + 648n^6 + 1296n^7)\\ (x,\,y) = (d_1,\;d_2) =\; &(12n^2 + 648n^6 + 1296n^7,\\ & 1 + 6n^2 + 36n^3 + 108n^4 + 2592n^7 + 3888n^8)\end{align}
Starting with the single polynomial $c_1$, we can again find six $c_k$. Likewise for the $d_k$. And we used $(c_1, c_2)$ to find $(d_1, d_2)$ by the process described in Ellipse 1. Incidentally, $$c_1^3+c_2^3+z^3 = (z+1)^3$$
where $z=12n^2 + 72n^4 + 1296n^6 + 3888n^8 + 46656n^{10},$ so this $z$ is immune to sign changes too.
Ellipse 3: Factoring $\,x^3+y^3+z^3 = (z+1)^3,\,$ where $z = (12n^2 + 324n^4)x + (1 + 12n^2 + 324n^4)(y - 1).$
Initial lattice points:
\begin{align}(x,\,y) = (e_1,\;e_2) =\; &(6n + 18n^2 + 180n^3 + 540n^4 + 3240n^5 + 5832n^6 + 34992n^7,\\ & 1 + 36n^2 - 72n^3 + 864n^4 - 3240n^5 + 5832n^6 - 34992n^7)\\ (x,\,y) = (f_1,\;f_2) =\; &(12n^2 - 72n^3 + 216n^4 - 3240n^5 + 5832n^6 - 34992n^7,\\ & 1 - 6n + 42n^2 - 252n^3 + 1404n^4 - 6480n^5 + 23328n^6 - 69984n^7 + 314928n^8)\end{align}
Yet again, starting with the single polynomial $e_1$, we can find six $e_k$. Likewise for the $f_k$. And we used $(e_1, e_2)$ to find $(f_1, f_2).$ Incidentally,
$$e_1^3+e_2^3+z^3 = (z+1)^3$$
where $z=36n^2 + 1512n^4 + 40176n^6 + 594864n^8 + 3779136n^{10}$, still immune to sign changes.
P.S. As to the question of how the pairs of initial polynomials $a_1,b_1,$ etc were originally found, it is a combination of intuition, laborious guesswork, and Mathematica's "InterpolatingPolynomial[] function". But it is nice to know we can derive one from the other.
P.S. The integer $6$ seems ubiquitous for these ellipses. Note the number N of Ellipse One's lattice points for $n=1,2,3,4\dots,36$ as calculated by the Alpertron calculator is N = (12, 48, 24, 36, 72, 24, 72, 48, 48, 96, 192, 48, 72, 288, 72, 96, 96, 96, 144, 96, 288, 24, 96, 192, 96, 144, 48, 288, 24, 384, 144, 48, 192, 48, 72, 48), all multiples of $12$.