A 4-variable version of the infamous ABC Conjecture says the following: Let $a,b,c,d\in\mathbb{Z}$ be non-zero, satisfy $a+b+c+d=0$ and $\gcd(a,b,c,d)=1$, and no subsum of two or three of $a,b,c,d$ equal to $0$. Then for every $\epsilon>0$ there is a constant $K_\epsilon$ such that
$$ \max\{|a|,|b|,|c|,|d|\} \le K_\epsilon \prod_{p\mid abcd} p^{1+\epsilon}. $$
Applying this to an expression of the form $a^n+b^n-c^n-k$ gives a very strong bound. Assuming that I haven't made an error (which is quite possible), I think that if $n\ge5$ (and assuming the ABCD conjecture), then for any $k$, the equation
$$ a^n + b^n - c^n = k $$
has only finitely many solutions $a,b,c\in\mathbb{Z}$ with $|a|,|b|,|c|$ distinct and non-zero.

Actually, I guess the same (more or less) should be true for $n=4$. The point is that the surface
$$ x^n+y^n-z^n=k $$
is of general type for $n\ge5$, so the Bombieri-Lang conjecture says that the solutions in *rational numbers* $(x,y,z)\in\mathbb{Q}^3$ are not Zariski dense (lie on a finite set of curves). This also follows from Vojta's conjecture. And for $n=4$, the equation defines an affine piece of a K3 surface, so Vojta's conjecture implies that the set of integer solutions likewise lies on a finite set of curves.

So your problem fits into a general framework, and for example, these statements are known if you replace $\mathbb Z$ by the ring of polynomials $\mathbb C[t]$. And as July suggests, you might want to read about how such problems are normally written, since your notation is not at all standard (and somewhat hard to parse).

non-trivialsolutions. Of course, then one needs to figure out what are the trivial cases. You've mentioned two of them. So let's insist that solutions have $a,b,c\ge2$ and $b\ne c$. $\endgroup$ – Joe Silverman Aug 10 '15 at 12:38