OK so I think that Jagy wants to make the following conjecture:

CONJECTURE 1: an integer $C$ is not representable by the form F(x,y,z)=2x^2+xy+3y^2+z^3-z if, and only if, $C$ is odd and $27C^2-4=23D^2$ with $D$ an integer.

I have a proof strategy for this. I am too lazy to fill in some of the details though, so maybe a bit of it doesn't work, but it should be OK. However, I am also reliant on a much easier-looking conjecture (which I've tested numerically so should be fine, but I can't see why it's true):

CONJECTURE 2: if $C$ is odd and $27C^2-4=23D^2$, then there's no prime p
dividing D of the form $2x^2+xy+3y^2$.

So I am claiming Conj 2 implies Conj 1. I don't know how to prove Conj 2
but it looks very accessible. Note that the Pell equation is related to units
in $\mathbf{Q}(\sqrt{69})$ and the $2x^2+xy+3y^2$ is related to factorization
in $\mathbf{Q}(\sqrt{-23})$. I've seen other results relating the arithmetic
of $\mathbf{Q}(\sqrt{D})$ and $\mathbf{Q}(\sqrt{-3D})$.

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Ok, so assuming Conjecture 2, let me sketch a proof of Conjecture 1.

The Pell equation is intimately related to the recurrence relation

$$t_{n+2}=25t_{n+1}-t_n$$

with various initial conditions. For example the positive $C$s which
are solutions to $27C^2-4=23D^2$ are all generated by this recurrence
starting at $C_1=C_2=1$, and the $D$s are all generated by the same
recurrence with $D_1=-1$ and $D_2=1$. Note that $C_n$ is even iff $n$
is a multiple of 3, and (by solving the recurrence explicitly) one
checks easily that $C_{3n}=(3C_{n+1})^3-(3C_{n+1})$, so we've represented
the even solutions to the Pell equation as values of $F$ (with $x=y=0$).

Let's then consider the odd solutions to the Pell equation. Say $C$
is one of these. We want to prove that there is no solution in
integers $x,y,z$ to 

$$2x^2+xy+3y^2=z^3-z+C.$$

Let's do it by contradiction. Consider the polynomial $Z^3-Z+C$. First
I claim it's irreducible. This is because it is monic, of degree 3,
and has no integer root, because $C$ is odd. Next I claim that
the splitting field contains $\mathbf{Q}(\sqrt{-23})$. This is
because of our Pell assumption and the fact that the discriminant
of $Z^3-Z+C$ is $4-27C^2$. Next I claim that the splitting
field of $Z^3-Z+C$ is in fact the Hilbert class field of
$\mathbf{Q}(\sqrt{-23})$. I only know an ugly way of seeing this:
if $\theta$ is a root of $Z^3-Z+1=0$ then I know recurrence relations
$e_n$, $f_n$ and $g_n$ (all defined using the relation above but with
different initial conditions) with $e_n\theta^2+f_n\theta+g_n$ a root of
$Z^3-Z+C_{3n+1}$, and other relations giving roots of $Z^3-Z+C_{3n+2}$
and $Z^3-Z-C_{3n+1}$ and $Z^3-Z-C_{3n+2}$. Most unenlightening but it
does the job because it embeds $\mathbf{Q}(\theta)$ into the splitting
field, and the Galois closure of $\mathbf{Q}(\theta)$ is the Hilbert
class field of $\mathbf{Q}(\sqrt{-23})$. 

Right, now for the contradiction, assuming Conjecture 2. Let's assume
that $C$ is a solution to the Pell, and $z^3-z+C$ can be written $2x^2+xy+3y^2$.
Now $C$ is odd so $z^3-z+C$ isn't zero, and hence it's positive,
so it's the norm of a non-principal ideal~$I$ in the integers $R$ of
$\mathbf{Q}(\sqrt{-23})$. This ideal $I$ is a product of prime ideals,
and $I$ isn't principal, so one of the prime ideals had better also not
be principal. Say this prime ideal has norm $p$. We conclude that $p$
divides $z^3-z+C$ and $p$ is of the form $2x^2+xy+3y^2$. Note in
particular that this implies $p\not=23$. Also $p\not=3$, because $C$
is odd and (because of general Pell stuff) hence prime to 3.

CASE 1: $p$ is coprime to $D^2$ (with $27C^2-4=23D^2$). In this
case the polynomial $Z^3-Z+C$ has non-zero discriminant mod $p$
(because $p\not=23$) and furthermore has a root $Z=z$ mod $p$.
Hence mod $p$ the polynomial either splits as the product of a linear
and a quadratic, or the product of three linears. This tells us
something about the factorization of $p$ in the splitting field
of $Z^3-Z+C$: either $p$ remains inert in $\mathbf{Q}(\sqrt{-23})$,
or it splits into 6 primes in the splitting field and hence splits
into two principal primes in $\mathbf{Q}(\sqrt{-23})$ (because the
principal primes are the ones that split completely in the Hilbert
class field). In either case $p$ can't be of the form $2x^2+xy+3y^2$,
so this case is done.

CASE 2: This is simply Conjecture 2.

In both cases we have our contradiction, and
so we have proved, so far, assuming Conjecture 2, that a solution $C$ to $27C^2-4=23D^2$
is representable as $2x^2+xy+3y^2+z^3-z$ iff it's even. 

Note that Conjecture 2 can be verified by computer for explicit values
of $C$, giving unconditional results---for example I checked in just
a few seconds that any odd $C$ with $|C|<10^{72}$ and satisfying the
Pell equation was not representable by the form, and that result
does not rely on anything. At least that's something concrete for Jagy.

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OK so what about the other way: say $27C^2-4$ is not 23 times a square.
How to go about representing $C$ by our form? Well, here I am going to
be much vaguer because there are issues I am simply too tired to deal
with (and note that this is not the question that Jagy asked anyway).
Here's the idea. Look at the proof of Theorem 2 in Jagy's pdf Mordell.pdf.
Here Mordell gives a general algorithm to represent certain integers
by (quadratic in two variables) + (cubic in one variable). If you
apply it not to the form we're interested in, but to the following
equation:

$$x^2+xy+6y^2=z^3-z+C$$

then, I didn't check all the details, but I convinced myself that they
could easily be checked if I had another hour or two, but I think that
the techniques show that whatever the value of $C$ is, this equation
has a solution. The idea is to fix $C$, let $\theta$ be a root of
the cubic on the right (which we can assume is irreducible, as if it
were reducible then we get a solution with $x=y=0$), to rewrite the right
hand side as $N_{F/\mathbf{Q}}(z-\theta)$, with $F=\mathbf{Q}(\theta)$
and now to try and write $z-\theta$ as $G^2+GH+2H^2$ with
$G,H\in\mathbf{Z}[\theta]$. Mordell does this explicitly (in a slightly
different case) in the pdf. The arguments come out the same though,
and we end up having to check that a certain cubic in four variables
has a solution modulo~23 with a certain property. I'll skip the painful
details. The cubic depends on $C$ mod 23, and so a computer calculation
can deal with all 23 cases.

Once this is done properly we have a solution to $x^2+xy+6y^2=z^3-z+C$,
so we have written $z^3-z+C$ as the norm of a principal ideal in
the integers of $\mathbf{Q}(\sqrt{-23})$. What we need to do now is
to write it as the norm of a non-principal ideal, and of course we'll
be able to do this if we can find some prime $p$ dividing $z^3-z+C$
which splits in $\mathbf{Q}(\sqrt{-23})$ into two non-principal
primes, because then we replace one of the prime divisors above $p$
in our ideal by the other one. What we need then is to show that
if the discriminant of $z^3-z+C$ is not $-23$ times a square,
then there _is_ some prime $p$ of the form $2x^2+xy+3y^2$ dividing
some number of the form $z^3-z+C$ which is the norm of a principal
ideal. This should follow from the Cebotarev density theorem, because
Mordell's methods construct a huge number of solutions to $x^2+xy+6y^2=z^3-z+C$
which are "only constrained modulo 23", and so one should presumably
be able to find a prime which splits in $\mathbf{Q}(\sqrt{-23})$,
splits completely in the splitting field of $z^3-z+C$ and doesn't
split completely in the splitting field of $z^3-z+1$. I have run out
of energy to deal with this point however, so again there is a hole here.
This issue seems analytic to me, and I am not much of an analytic guy.