On the mixed sum of three k-th powers Let the set $S_k=\{\pm x^k \pm y^k \pm z^k \ \vert \ x,y,z \in \mathbb{Z} \}$.
Note that the signs are independently positive or negative.  
First of all $S_2 = \mathbb{Z}$ because (see the answers of this post):
$2m = (m+1)^2 - m^2 - 1$ and $2m+1 = (m+3)^2 - (m+2)^2 - 4$
It was proved by congruence computation that  $n \in S_3$ implies $n \not \equiv 4,5  \pmod{9} $, and the converse was conjectured and checked for $n≤1000$ except $33$, $42$, $74$, $114$, $165$, $390$, $579$, $627$, $633$, $732$, $795$, $906$, $921$, and $975$ (see this paper) (see also here).
Question: Is there a similar conjecture and computational results for some small $k>3$?   
We find by congruence computation that: 


*

*$n \in S_4$ implies $n \not \equiv  4,5,6  \pmod{8} $ 

*$n \in S_5$ implies $n \not \equiv  4,5,6,7  \pmod{11} $ 

*$n \in S_6$ implies
$n \not \equiv   4,5  \pmod{7} $
$n \not \equiv   4,5,6  \pmod{8} $
$n \not \equiv  4,5,6,7 \pmod{9} $
$n \not \equiv  4, 5, 6, 7, 8, 9 \pmod{13} $

*$n \in S_7$ implies
$n \not \equiv   8, 9, 20, 21  \pmod{29} $
$n \not \equiv   10, 16, 17, 26, 27, 33  \pmod{43} $
$n \not \equiv  4, 9, 15, 22, 23, 24, 25, 26, 27, 34, 40, 45 \pmod{49} $


and  the converse of each point could be conjectured. 
Remark: this answer states that assuming the generalized abc conjecture, $S_k$ should have zero density for $k>18$, which prevents, in this case, a conjecture describing this set by finitely many congruences, as above.
 A: EDIT This answers previous revision, quite different question.

Let $k = 5$. I think a single representation outside
your forbidden congruences would violate "Vojta's more general abc conjecture".
In A more general abc conjecture, p. 7 Paul Vojta conjectures:

If $x_0,\ldots x_{n-1}$ are nonzero coprime integers satisfing $x_0 + \cdots x_{n-1}=0$
$$ \max\{|x_0|,\ldots |x_{n-1}|\} \le C \prod_{p\mid x_0 \cdots x_{n-1}}p^{1+\epsilon}\qquad (1) $$
for all $x_0 , \ldots, x_{n-1}$ as above outside a proper Zariski-closed subset.

Let $n=2^m 11^l$, where $m,l$ are coprime to $5$ and $m,l$ are coprime.
$m,l$ can be arbitrary large. The radical of $n$ is constant
and we have $n=2^m 11^l=x^5+y^5+z^5$.
By Vojta's conjecture, as $m,l$ vary as above, every single solution
must be on proper Zariski-closed subset, which appears highly unlikely to
me.
To ensure coprimality, clear the gcd.
$2$ can be replaced by any other positive natural
and $n$ large, but with small radical satisfying your congruences
will in general work, unless clearing the gcd solves it.

Similar argument for larger $k$ and $n$ sufficiently larger
than its radical would contradict also the n-conjecture.
The n-conjecture
is a generalization of abc and basically says that the if
$a_1 + \ldots + a_n=0$, no proper subsum vanishes and $a_i$
are coprime, then the radical of $a_1\cdots a_n$ can't be
too small (without exceptional set).
