This goes back to my first MO question https://mathoverflow.net/questions/12486/integers-not-represented-by-2-x2-x-y-3-y2-z3-z and before that to Irving Kaplansky.
I put a fair amount of related information at http://zakuski.math.utsa.edu/~jagy/inhom.cgi
My original construction was inhomogeneous polynomials of the same sort as $2 x^2 + xy + 3 y^2 + z^3 - z.$ The result is that $2 x^2 + xy + 3 y^2 + z^3 - z \neq \pm C,$ where $C = 1, 599, 14951, 9314449, $ and in general Pell type $27C^2 - 23 F^2 = 4$ for some integer $F.$
Over the weekend I shared this with a conference and gave a vague mention of the case when the binary quadratic form is indefinite. The example I mentioned (disc 229) was $3 x^2 + 13 xy - 5 y^2 + z^3 - 4z .$ By my computer search, it appears that $3 x^2 + 13 xy - 5 y^2 + z^3 - 4z \neq \pm 1 .$
The very next example is discriminant 257. In Henri Cohen's A Course in Computational Number Theory the cubic, table B.4, page 522, is given as $x^3 - 5 x - 3.$ I like this version, the computer gets $2 x^2 + 15 xy - 4 y^2 + z^3 - 5z \neq \pm 3 .$ This includes facts such as : $2 x^2 + 15 xy - 4 y^2$ represents a number $n$ if and only if it represents $-n.$
I was looking for larger discriminants at TABLE and found $x^3 - x^2 - 4x +3$ instead. It does have discriminant $257,$ same Galois group. John Voight gives the same. However, I now find $2 x^2 + 15 xy - 4 y^2 + z^3 - z^2 - 4z \neq - 3 .$ It gives the value $3$ when, for example, $x=3, y=-3, z=6$ or $x=5, y=-1, z=4$
Oh, the pattern suggests a finite count of missing numbers when the binary is indefinite. I'll need to find that again. Found it, for Cohen's version, $\Delta(x^3 - 5x - c) = 500 - 27 c^2$ and $\Delta = 257 $ means $c = \pm 3$
So, there is the question, why are the given cubics different, $x^3 -5x -3$ and $x^3 - x^2 - 4x +3;$ also why the seeming difference in the missing integers, two for the first version, single missing number in the second?
From comments by Aphelli confirming they are the same cubic field, $$ \left( 3 - x^2 \right)^3 - 5 \left( 3 - x^2 \right) - 3 = - \left( x^3 - x^2 - 4x + 3 \right) \left( x^3 + x^2 - 4x - 3 \right) $$
$$ \left( x^2 - x - 3 \right)^3 - \left( x^2 - x - 3 \right)^2 -4 \left( x^2 - x - 3 \right) +3 = \left( x^3 - 5x - 3 \right) \left( x^3 - 3x^2 - 2x + 7 \right) $$ In particular, when the factor of interest on the right hand side is zero, the left hand side is also zero.