EDIT: Hendrik Lenstra emailed me a proof of Conjecture 2. I'll append it below. So Jagy's question is now solved.
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 $\mathbf{Q}(\sqrt{-3D})$.
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.
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.
EDIT: OK so here, verbatim, is an email from Lenstra in which he establishes Conjecture 2. Note: when he says "2)" he means the statement that if C is an odd solution to the Pell and if theta is a root of $X^3-X+1=0$ then $\mathbf{Z}[\theta]$ contains a root of $X^3-X+C$ (I sketched a proof of this above).
Let eta in Z[theta] be a zero of X^3-X+C. Then |D| = index(Z[theta]:Z[eta]). Now let p be a prime number of the form 2x^2+xy+3y^2. To be proved: p does not divide D. By class field theory over Q(sqrt(-23)), the prime p is inert in Z[theta], so Z[theta]/pZ[theta] is a field of size ppp. Look at the image e (say) of eta in that field. It is nonzero, since from eta's equation it is clear that eta/p is not an algebraic integer. First case: e is in the prime field. Then since the trace of eta is 0, the trace 3e of e is zero but e isn't, so p = 3. And 3 visibly doesn't divide D by 27CC-4=23DD. (Actually, if we believe 2), then this first case cannot occur, since e in prime field implies Z[eta] subset Z + pZ[theta] hence 3|index = |D| after all.) Second case: e is not in the prime field. Then e generates the entire field, so Z[theta] = Z[eta] + pZ[theta], so p acts surjectively on the finite abelian group Z[theta]/Z[eta], so its order |D| is not divisible by p. End of proof.