Coprime integer solutions to $x_1^{r-1} y_1^r + x_2^{r-1} y_2^r = x_3^{r-1} y_3^r$

Question

The question asks what is known about integer solutions $(\mathbf{x}, \mathbf{y}) = ((x_1, x_2, x_3), (y_1, y_2, y_3))$ to the equation

$$\displaystyle x_1^{r-1} y_1^r + x_2^{r-1} y_2^r = x_3^{r-1} y_3^r$$

where $r \geq 4$ is an integer, $x_1 x_2 x_3 y_1 y_2 y_3 \ne 0$, and $\gcd(x_1 y_1, x_2 y_2, x_3 y_3) = 1$. In particular, does one expect infinitely many solutions, and can one prove an asymptotic lower bound?

When $r = 2,3$ this problem has been studied by Le Boudec and Browning/Valckenborgh respectively, which answers the question in the affirmative in these cases. Le Boudec establishes the exact asymptotic order of magnitude, while Browning-Valckenborgh establishes a lower bound which is expected to be the correct order of magnitude and an upper bound of $O_\epsilon(T^{3/5 + \epsilon})$.

As Henri Cohen remarks below, when $r = 4$ it is possible to choose $y_1 = a, y_2 = b, y_3 = c$ in such a way that the genus one plane cubic curve $a^4 x_1^3 + b^4 x_2^3 = c^4 x_3^3$ has a rational point and has positive rank as an elliptic curve (in particular, $y_1 = y_2 = 1, y_3 = 9$ gives one such choice). This trick will not work when $r$ is sufficiently large.

Answer 1

Assuming the abc conjecture, there are only finitely many solutions with $r\geq 5$. Indeed, more generally, abc conjecture implies there are only finitely many sums of the form $a+b=c$ in which $a,b,c$ are coprime and all prime exponents in them are at least $4$. Indeed, for any such triple we have $$rad(abc)\leq (abc)^{1/4}\leq c^{3/4}$$ (assuming $a,b,c$ are positive, if not you can switch signs and permute them.) Therefore one definitely shouldn't expect there to be infinitely many solutions for $r\geq 5$. However I doubt an unconditional proof is known, even for a fixed value of $r$.