To be precise, I am asking:

**Does there exist an integer $k$ such that there do not exist (possibly negative) integers $x,y,z$ satisfying $x^4+y^4=z^3+k$?**

Heuristically the answer must be yes, in fact, one expects that almost every $k$ should work (this is just because the sum of the reciprocals of the exponents is $\frac13+\frac14+\frac14 = \frac56<1$, so one ought to expect that up to a large bound $N$, something like $N^{5/6}$ values of $k$ are representable). For all I know, $-2$ and $4$ might already be two examples of numbers which can't be represented as sums of two fourth powers minus a cube, but I haven't been able to prove this.

What I can prove is that there are *no local obstructions*: for any integers $n,k$ with $n\ne 0$, we can find integers $x,y,z$ with $(x,y,z)=1$ such that $x^4+y^4\equiv z^3+k\pmod{n}.$ Using the Chinese Remainder Theorem and Hensel's Lemma one can quickly reduce this claim to the case that $n=p$ is a prime. The most interesting case is when $p\equiv 1\pmod{12}$ and $p\nmid k$, and in this case we can use a trick similar to the proof of Chevalley-Warning to count the number of solutions $N_p$ modulo $p$. We start with the easy congruence

$N_p \equiv \sum_{x,y,z} (1-(x^4+y^4-z^3-k)^{p-1}) \equiv -\sum_{x,y,z}\sum_{a+b+c+d=p-1} \frac{(p-1)!}{a!b!c!d!} x^{4a}y^{4b}(-z)^{3c}(-k)^d\pmod{p},$

and then note that if we fix $(a,b,c,d)$ and sum over $x,y,z$, we can only get a nonzero contribution when $p-1\mid 4a,4b,3c$ and $a,b,c>0$. From this we see that

$N_p \equiv \frac{(p-1)!}{\left(\frac{p-1}{4}\right)!\left(\frac{p-1}{4}\right)!\left(\frac{p-1}{3}\right)!\left(\frac{p-1}{6}\right)!} (-k)^{\frac{p-1}{6}}\not\equiv 0\pmod{p},$

so in particular $N_p \ne 0$.