Question: Do there exist integers $(x,y,z)\neq (0,0,0)$ such that
$$
13x^4+11y^4=8z^4 ?
$$


Some motivation: This is currently the smallest (in a sense defined here [On the smallest open Diophantine equations: beyond Hilbert's 10 problem](https://mathoverflow.net/questions/411958)) three-monomial homogeneous equation that I do not see how to solve. Search returns no non-trivial solutions. Magma says that the equation is locally solvable. Another standard method to solve equations of the form $ax^4+by^4+cz^4=0$ is to look at genus 1 curves $ax^2+by^4+cz^4=0$, $ax^4+by^2+cz^4=0$ and $ax^4+by^4+cz^2=0$, and if any one of them has rank $0$, the rest is easy. But in this example the ranks are $1,1,2$.


I found many papers studying equations $x^4+y^4=cz^4$ systematically for all $c$ up to $10,000$, but cannot find a similar systematic study of more general $ax^4+by^4+cz^4=0$.



Existence of rational points on a generalized Fermat quartic
This question is amenable to the use of the Mordell-Weil sieve. (For a good introduction to this technique, see the paper [here](https://arxiv.org/abs/0906.1934).) In this situation, there is a simple version of it (using a single prime) suffices.


Let $C : 13x^{4} + 11y^{4} = 8z^{4}$, and let $E : y^{2} = x^{3} + 3146x$. (This elliptic curve is isomorphic to $13x^{4} + 11y^{2} = 8$. I chose this elliptic curve because it has rank $1$.) Therefore, we have a map $\phi : C \to E$. For a prime $p$ for which $C$, $E$ and the map $\phi$ have good reduction, we obtain the following commutative diagram.
$\require{AMScd}$
\begin{CD}
C(\mathbb{Q}) @>\phi>> E(\mathbb{Q})\\
@VV r V @VV \beta V\\
C(\mathbb{F}\_{p}) @>\alpha>> E(\mathbb{F}\_{p})
\end{CD}
The vertical maps are reduction modulo $p$, and the horizontal maps are the map from $C$ to $E$ (the top row in characteristic zero, the bottom row in characteristic $p$).


While we don't know $C(\mathbb{Q})$, the other three sets are computable. (In particular, $E(\mathbb{Q}) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}$ and is generated by $(0,0)$ and $(3041064/87025,10028940012/25672375)$.)


In the event that there is a rational point $Q$ in $C(\mathbb{Q})$, we have that $(\alpha \circ r)(Q) = (\beta \circ \phi)(Q)$ is in ${\rm im~} \alpha \cap {\rm im~} \beta$. So if ${\rm im~} \alpha \cap {\rm im~} \beta = \emptyset$, it follows that $C(\mathbb{Q})$ is empty. Computer calculations show that ${\rm im~} \alpha \cap {\rm im~} \beta = \emptyset$ for $p \in \{ 17, 197, 1321, 2089 \}$. It follows that $C(\mathbb{Q}) \subseteq \mathbb{P}^{2}(\mathbb{Q})$ is empty.


Even if a single prime didn't suffice, it would also be possible to combine congruence information from different primes to try to obtain a contradiction.


