As I understand the question, the OP wishes to solve,
$$x_1^k + x_2^k = x_3^k + x_4^k\tag1$$
for $k = 5$ and where the $x_i$ are roots of quartics. If we allow that the $x_i$ are roots of different quartics, then there are in fact solutions for $k = 5,6,8$.
I. k = 5
The first identity below just uses quadratics,
$$(a\sqrt2+b)^5+(-b+c\sqrt{-2})^5 = (a\sqrt2-b)^5+(b+c\sqrt{-2})^5$$
with Pythagorean triples $(a,b,c)$ and was known by Desboves. The second is by yours truly,
$$(\sqrt{p}+\sqrt{q})^5+(\sqrt{p}-\sqrt{q})^5 = (\sqrt{r}+\sqrt{s})^5 + (\sqrt{r}-\sqrt{s})^5$$
where,
\begin{align} p &= 5vw^2,\quad q = -1+uw^2\\ r &= 5v,\quad\quad s = -(u+10v)+w^3\end{align}
and $w = u^2+10uv+5v^2$.
II. k = 6
We use the same form,
$$(\sqrt{p}+\sqrt{q})^6+(\sqrt{p}-\sqrt{q})^6 = (\sqrt{r}+\sqrt{s})^6 + (\sqrt{r}-\sqrt{s})^6$$
where,
\begin{align} p &= -(a^2+14ab+b^2)^2+(ac+bc+13ad+bd)(c^2+14cd+d^2)\\ q &= \;\;(a^2+14ab+b^2)^2-(ac+13bc+ad+bd)(c^2+14cd+d^2)\\ r &= \;\;(c^2+14cd+d^2)^2-(ac+13bc+ad+bd)(a^2+14ab+b^2)\\ s &= -(c^2+14cd+d^2)^2+(ac+13bc+ad+bd)(a^2+14ab+b^2) \end{align}\begin{align} p &= -(a^2+14ab+b^2)^2+(ac+bc+13ad+bd)(c^2+14cd+d^2)\\ q &= \;\;(a^2+14ab+b^2)^2-(ac+13bc+ad+bd)(c^2+14cd+d^2)\\ r &= \;\;(c^2+14cd+d^2)^2-(ac+13bc+ad+bd)(a^2+14ab+b^2)\\ s &= -(c^2+14cd+d^2)^2+(ac+bc+13ad+bd)(a^2+14ab+b^2) \end{align}
III. k = 8
Still using the same form,
$$(\sqrt{p}+\sqrt{q})^8+(\sqrt{p}-\sqrt{q})^8 = (\sqrt{r}+\sqrt{s})^8 + (\sqrt{r}-\sqrt{s})^8$$
where, \begin{align} p &= n^3-2n+1\\ q &= n^3+2n-1\\ r &= n^3-n-1\\ s &= n^3-n+1\end{align}
P.S. Unfortunately, $k= 7$ and $k=9$ does not seem to be amenable to the same approach.