It has been conjectured by Euler, Ekl, etc, that $$x_1^k+x_2^k=x_3^k+x_4^k$$ has no solution for $k>4$ if the $x_i$ are algebraic numbers of deg $1$.
However, if these are deg $2$, then there are solutions when $k>4$. A parametrization by Desboves is if $a^2+b^2=c^2$, then,
$$(a\sqrt2+b)^5+(-b+c\sqrt{-2})^5=(a\sqrt2-b)^5+(b+c\sqrt{-2})^5$$
Update: There is even for $k=11$, $$(1+\sqrt{-7})^{11}+(1-\sqrt{-7})^{11}=(1+\sqrt{-11})^{11}+(1-\sqrt{-11})^{11}$$ which answers the title of the post, but leaves the question of how high $k$ can get.
Q: But if we now consider the $x_i$ to be algebraic numbers of deg $4$ of form, $$(\sqrt{p}+\sqrt{q})^k+(\sqrt{p}-\sqrt{q})^k= (\sqrt{r}+\sqrt{s})^k+(\sqrt{r}-\sqrt{s})^k\tag1$$ where $p,q,r,s$ are integers, does it have non-trivial solutions for $\color{blue}{k>11}$?
It has parametric ones for $k\leq10$. (See this answer for $k\leq6$.) Also,
For $k=7,9:$ $$ (\sqrt{a+c}+\sqrt{a+2b-c})^k+(\sqrt{a+c}-\sqrt{a+2b-c})^k = \\(\sqrt{a+c}+\sqrt{a-2b-c})^k+(\sqrt{a+c}-\sqrt{a-2b-c})^k$$ $$\begin{aligned} &\text{If}\;k=7,\; \text{then}\; 4a^2+b^2=c^2.\\ &\text{If}\;k=9,\; \text{then}\; 6 a(a^2 + 5 b^2) + 2 b^2 c - 3 a c^2 = 0. \end{aligned}$$
For $k=8,10:$ $$ (\sqrt{2a+c}+\sqrt{-2a+c})^k + (\sqrt{2a+c}-\sqrt{-2a+c})^k = \\(\sqrt{2b+c}+\sqrt{-2b+c})^k + (\sqrt{2b+c}-\sqrt{-2b+c})^k$$ $$\begin{aligned} &\text{If}\;k=8,\; \text{then}\; a^2+b^2=2c^2.\\ &\text{If}\;k=10,\; \text{then}\; a^2+b^2=c^2. \end{aligned}$$
which are just quadratics in $c$ and easily solvable. But extending these for $k=11,12$, respectively, yield quartics in $c$ and are harder. (And these special cases seem to always involve a negative integer under the square root.)
P.S. A different parameterization for $k=8$ involves, of all things, the minpoly of the golden ratio $\phi$ and the plastic constant $P$.
Q: In summary, does $(1)$ have non-trivial solutions for $k>11$ as well?
(I have done a computer search, but my search radius may have been too small.)
