Can $9xy$ divide $1+x^2+x^3+y^2$ for integers $x,y$? Equivalently, do there exist integers $x,y,z$ such that $$ 1 + x^2 + x^3 + y^2 + 9 x y z = 0 \quad ? $$ This equation arises in the search for the shortest cubic equation with open Hilbert 10-th problem, see e.g. On the shortest open cubic equation
I searched for $|y|$ up to 100 millions and $|x|$ up to 300 millions, with no solutions found.
It is easy to see that $y \equiv 3 \bmod 6$, while $x \equiv 4 \bmod 9$ and is not divisible by $4$.
The equation is solvable in integers. Take, for example, $$ x = -19578556686240310295378317903565, \\ y = -101658411567714319887, \\ z = 418962851513108789978912616277180591709694. $$
Verification can be done by substitution.
I found this solution by using the transformations described by Denis Shatrov in the comment. We consider equation $$ 1 + x^2 + x^3 + y^2 + x y z = 0 \quad\quad (1) $$ and look for a solution such that $z$ is divisible by $9$. If we start with any solution $(x_0,y_0,z_0)$, then, as observed by Denis, $$ (x_1,y_1,z_1)=\left(\frac{y_0^2+1}{x_0}, y_0, -\frac{1+x_1+x_1^3+y_1^2}{x_1y_1}\right) $$ solves equation $$ 1 + x + x^3 + y^2 + x y z = 0 \quad\quad (2). $$ Then $$ (x_2,y_2,z_2)=\left(x_1, \frac{x_1^3+x_1+1}{y_1}, -\frac{1+x_2+x_2^3+y_2^2}{x_2y_2}\right) $$ is also a solution to (2), while $$ (x_3,y_3,z_3)=\left(\frac{y_2^2+1}{x_2}, y_2, -\frac{1+x_3^2+x_3^3+y_3^2}{x_3y_3}\right) $$ is again a solution to (1).
By doing modulo 9 analysis, I observed that if $(x_0,y_0,z_0)$ is $(4,0,3)$ modulo $9$, then $(x_3,y_3,z_3)$ is $(4,6,0)$ modulo $9$. An easy computer search returned solution $(x_0,y_0,z_0) = (-3965, 1446687, 354)$ to (1) which is $(4,0,3)$ modulo $9$. Then the corresponding $(x_3,y_3,z_3)$ is a solution to (1) with $z$ divisible by (9), hence $(x_3,y_3,z_3/9)$ is an integer solution to the original equation $1 + x^2 + x^3 + y^2 + 9x y z = 0$. As mentioned above, its correctness can be easily verified by direct substitution.
