Fourteen years ago, in 1999 (has it been that long?) Merignac started a search for,
$$x_1^6+x_2^6+x_3^6+\dots+x_7^6 = \color{red}z^6$$
with the hope that one $x_i =0 $ using the five congruence classes,
$$\begin{aligned} &42^6(x_1^6+x_2^6+\dots+x_5^6)+(42x_6)^6+(1x_7)^6 = z^6\\ &42^6(x_1^6+x_2^6+\dots+x_5^6)+(21x_6)^6+(2x_7)^6 = z^6\\ &42^6(x_1^6+x_2^6+\dots+x_5^6)+(14x_6)^6+(3x_7)^6 = z^6\\ &42^6(x_1^6+x_2^6+\dots+x_5^6)+\,(7x_6)^6\,+\,(6x_7)^6 = z^6\\ &42^6(x_1^6+x_2^6+\dots+x_4^6)+(21x_5)^6+(14x_6)^6+(6x_7)^6 = z^6 \end{aligned}$$
So the first six $x_i$ are multiples of $7$, and primitive integer solutions are known for all five classes. The smallest known (found around 2000) by Meyrignac and Wannes de Smedt belongs to the 4th class,
$$42^6(195^6 + 260^6 + 440^6 + 506^6 + 580^6) + (7\times2747)^6 + (6\times5559)^6 = 34781^6$$
But it seems none has one $x_i = 0$ for $\color{red}z<730000$. (See Further work section of 2002 paper The Smallest Solutions to the Diophantine Equation $a^6+b^6+c^6+d^6+e^6=x^6+y^6$.)
To compare to 4th powers, one can primitively solve,
$$a^4(x_1^4+x_2^4+x_3^4)+x_4^4 = z^4$$
for $a=10$ or $a=20$ with the smallest being,
$$10^4(24^4+34^4+43^4)+599^4 = 651^4$$ $$20^4(19^4+83^4+94^4)+4907^4 = 4949^4$$
Question: Is $b=42$ overly restrictive? Can it be reduced to just $b = 21$? (As the fourth power example shows, $a = 10$ has a smaller solution. If so, maybe they overshot a solution in the $z<730000$ range?)
The restriction to $b = 42$ is correct because any solution of the Diophantine equation
$$x_1^6+x_2^6+\dots+x_6^6 = z^6$$
would have to meet constraints associated with each of the prime factors of 42:
a) If $a \not\equiv 0 \pmod{2}$ then $a^6 \equiv 1 \pmod{8}$;
b) If $a \not\equiv 0 \pmod{3}$ then $a^6 \equiv 1 \pmod{9}$;
c) If $a \not\equiv 0 \pmod{7}$ then $a^6 \equiv 1 \pmod{7}$.
(These can be found on this page of Meyrignac's website, albeit applied to a different 6th power equation.) To meet these constraints, a (primitive) solution must have $x_i \equiv 0 \pmod{2}$ for 5 of the 6 left hand terms, and similarly for mod 3 and mod 7. Hence a solution must have $x_i^6 \equiv 0 \pmod{42}$ for at least 3 of the 6 left hand terms.
There is also a restriction associated with 13 arising from the fact that:
d) If $a \not\equiv 0 \pmod{13}$ then $a^6 \equiv 1$ or $-1 \pmod{13}$.
This implies that a solution must have an odd number of terms (which could include $z$) congruent to 0 mod 13 and, among the remaining terms, an appropriate balance of sixth powers congruent to 1 and to -1 mod 13. To combine this restriction with those relating to 2, 3 and 7 would be quite tedious but it can be said that the five congruence classes are not restrictive enough to exclude all cases which are impossible for reasons of congruence.