It's quite easy to find integer solutions to,
$$x_0^k + x_1^k + \dots +x_9^k = 0$$
for $k = 1, 3, 5, 7$. One I found is, if $x^2-10y^2 = 9$, then,
$$1 + 5^k + (3+2y)^k + (3-2y)^k + (-3+3y)^k + (-3-3y)^k = (-2+x)^k + (-2-x)^k + (5-y)^k + (5+y)^k$$
However, Letac found in 1942 two special cases with $x_0 = 0$ (the case $x_0 = x_1 = 0$ is impossible), namely,
$$\{\,0, 34, 58, 82, 98\,\} = \{\,13, 16, 69, 75, 99\,\}$$
$$\{\,0, 63, 119, 161, 169\,\} = \{\,8, 50, 132, 148, 174\,\}$$
How did he do it?
The reference is A. Letac, Gazeta Matematica, 48 (1942), p. 68-69, and I'm hoping someone with JSTOR can explicitly give the answer.
Of the two solutions you mention, only the second appears in Letac’s note.
He used (among other things) the following two lemmas :
Fact 1. If $a_1^k+a_2^k+a_3^k=b_1^k+b_2^k+b_3^k$ for $k=2$ and $k=4$, then $$ a_1^p+(\pm a_2)^p+(\pm a_3)^p+(2b_1)^p+(2b_2)^p+(2b_3)^p= b_1^p+(\pm b_2)^p+(\pm b_3)^p+(2a_1)^p+(2a_2)^p+(2a_3)^p \ (p=2,4,6,8) $$
Fact 2. Under the hypotheses of fact 1, for $h=\frac{a_1+a_2-a_3}{4}$ we have
$$ (2a_1-3h)^k+(2a_2-h)^k+(2a_2-3h)^k+(2a_2-h)^k+(2b_3+h)^k= (2a_3+h)^k+(2a_3+h)^k+(2b_1-h)^k+(2b_1-h)^k+(2b_2-h)^k+(3h)^k \ (k=1,3,5) $$