This is related to my first MO question and Kevin Buzzard's conjecture at Integers not represented by $ 2 x^2 + x y + 3 y^2 + z^3 - z $

In December 2010 my question appeared in the M.A.A. Monthly, show that $4 x^2 + 2 x y + 7 y^2 - z^3 \neq \pm 2 m^3, \; \pm 32 m^3$ when $m$ has certain prime factorizations. The answers were due by April 2011 so I feel willing to mention it, although the answers have not appeared yet, I just got the August-September issue. Sigh.

A couple of days ago I thought I might check for identities, and found several good ones, showing that all odd numbers are represented for example. I believe there is no chance of completing this problem by identities owing to the non-represented numbers. So, that is the question, can anyone prove that $4 x^2 + 2 x y + 7 y^2 - z^3$ integrally represents everything else?

For verisimilitude, we have:

$$ \begin{array}{cc} x = 4 n^3 - 18 n^2 + 3 n - 21, & y = -16 n^3 - 18 n + 1, \\\ z = 12 n^2 + 12, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 6n+1 \end{array} $$

$$ \begin{array}{cc} x = 4 n^3 - 42 n^2 - 73 n - 359, & y = -16 n^3 - 48 n^2 - 146n - 111, \\\ z = 12 n^2 + 24n+ 88, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 6n-3 \end{array} $$

$$ \begin{array}{cc} x = 4 n^3 + 42 n^2 - 65 n + 417, & y = -16 n^3 + 48 n^2 - 166n + 137, \\\ z = 12 n^2 - 24n+ 98, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 6n+5 \end{array} $$

$$ \begin{array}{cc} x = 16 n^3 - 12 n^2 + 23 n + 6, & y = 8 n^3 - 24 n^2 + 28n - 27, \\\ z = 12 n^2 - 12 n+ 17, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 18n+10 \end{array} $$

$$ \begin{array}{cc} x = 16 n^3 - 12 n^2 + 3 n + 1, & y = 8 n^3 - 24 n^2 + 18n - 7, \\\ z = 12 n^2 - 12 n+ 7, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 18n-10 \end{array} $$

$$ \begin{array}{cc} x = 72 n^3 + 60 n^2 + 13 n, & y = -72 n^3 - 24 n^2 + 2 n + 1, \\\ z = 36 n^2 + 12 n+ 1, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 18n + 6 \end{array} $$

$$ \begin{array}{cc} x = 4 n^3 + 36 n^2 + 18 n + 135, & y = -16 n^3 - 60 n + 4, \\\ z = 12 n^2 + 42, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 24n + 4 \end{array} $$

$$ \begin{array}{cc} x = 9 n^3 - 30 n^2 + 29 n - 16, & y = -9 n^3 + 12 n^2 - 8 n + 2, \\\ z = 9 n^2 -12 n + 10, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 36n - 12 \end{array} $$

$$ \begin{array}{cc} x = 16 n^3 - 12 n^2 + 33 n + 7, & y = 8 n^3 - 24 n^2 + 30 n - 37, \\\ z = 12 n^2 -12 n + 21, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 162 n \end{array} $$

Furthermore, if we have a prime $q = 4 u^2 + 2 u v + 7 v^2,$ the fact that $h(-108) = 3$ and $2^2 + 27 \cdot 1^2 = 31$ shows that $ 4 x^2 + 2 x y + 7 y^2$ represents $q^3, \; 31 q^3, \; 25 q^3.$ As a result $q = 4 u^2 + 2 u v + 7 v^2 - z^3$ represents $2 q^3 = q^3 + q^3, \; 32 q^3 = 31 q^3 + q^3, \; -2 q^3 = 25 q^3 - 27 q^3. $ I'm not sure how to do $-32 q^3.$