I have managed to show that $x$ must be a multiple of $3$ and $y$ must be even, which produces the equation $$3^{10}s^{11}+2^{10}t^{11}=z^{11},$$ with $s=x/3$ and $t=y/2$. I have tried to approach this by applying Fermat's Little Theorem mod $11$ setting the tenth powers to $1$ but I've not managed to get any further.
Any help is really appreciated, thanks!
Simply take modulo $23$. If $23 \nmid n$, then: $$n^{22} \equiv 1 \pmod{23} \implies n^{11} \equiv \pm 1 \pmod{23}$$ Thus, for all $n \in \mathbb{N}$, we have $n^{11} \equiv -1,0,1 \pmod{23}$. Try substituting these values in your equation. You would see that the only possibility is $x^{11} \equiv y^{11} \equiv z^{11} \equiv 0\pmod{23}$. Substituting $x=23x_1$, $y=23y_1$ and $z=23z_1$, we get: $$2(x_1)^{11}+3(y_1)^{11} = 6(z_1)^{11}$$ Now, you will get $23$ divides $x_1,y_1,z_1$, which will give a process of infinite descent. This is clearly a contradiction. Thus, your equation does not have solutions in positive integers.