(This was posted previously in MSE without getting any answers.)

It is known that given primitive (co-prime) integer solutions to,

$$x_1^4+x_2^4+x_3^4+x_4^4 = z^4$$

then there is one $x_i$ such that $z^4-x_i^4$ is divisible by $d_4=5^4$. Additionally, Ward showed that if one of the $x_i$ is zero, then there is the further constraint that $z\pm x_j$ is divisible by $w_4=2^{10}$. For example,

$$673865^4+ 1390400^4+ 2767624^4 = 2813001^4$$

and,

$$z+x_3 = 2813001 + 2767624 = 5^4\cdot 8929$$

$$z-x_1 = 2813001 - 673865= 2^{10}\cdot 2089$$

Theorem: In general, for $k$ $k$th powers and co-prime terms,

$$x_1^k+x_2^k+x_3^k+\dots+x_k^k = z^k\tag1$$

if $k+1$ is prime, then there is one $x_i$ such that $z^k-x_i^k$ is divisible by $d_k = (k+1)^k$.

For $d_4 = 5^4 = 625$, this implies the smallest solution will have a term $>d_4/2 = \lfloor312\rfloor$. In fact,

$$30^4+120^4+315^4+\color{brown}{272}^4 = \color{brown}{353}^4$$

and $x_3+z =272+353=5^4.^{\color{brown}{Note}}$

For $d_6 = 7^6 = 117649$, the smallest solution (not yet found as of 2015) will have a term $>d_6/2 = \lfloor58824\rfloor$ hence will be relatively large and somehow "explains" why $k=8$ was found first,

$$90^8 + 223^8 + 478^8 + 524^8 + 748^8 + 1088^8 + 1190^8 + 1324^8 = 1409^8$$

since $8+1$ is composite and doesn't have the divisibility constraint $d_k$.

Question: Given $(1)$ where one of the $x_i$ is zero, is there an analogue to Ward's $w_k$ for $k=6$? If there is, what is it for general $k$?

$\color{brown}{Note}$: This has an analogue for $6$th powers when using seven addends,

$$1344^6+ 23268^6+ 25263^6+ 39088^6+ 48090^6+ 54138^6+ \color{brown}{54018}^6 = \color{brown}{63631}^6$$

and $x_7+z =54018+63631=7^6$.