Section 8.3.2 of recent book [1] studies the following problem. Define the length of a Polynomial Diophantine equation as the sum of degrees of monomials plus sum of base 2 logarithms of the coefficients, order the equations by length, and investigate the existence of integer solutions (you do not need to find all solutions). The shortest cubic equation left open in the book is $$ 1 + 3 x^3 + x y^2 + 6 y z^2 = 0 \quad\quad\quad (1) $$ The questions is where there exists integers $x,y,z$ satisfying this equation. With $t=xy$, this reduces to the solvability of $x+3x^4+t^2+6tz^2=0$ such that $x$ is a divisor of $t$.

Update 20.09.2024: Dmitry Ezhov now found some solutions to (1) starting from $(x,y,z) =(-1017461719,95574914,2350866170)$, see comments. So, this equation is now resolved. The next-shortest open cubic equation in [1] is
$$ y^2+10xyz+x^3-x-2=0. \quad\quad\quad (2) $$ Modulo $20$ analysis shows that $x=2$ mod $20$. After replacing $y$ with $-y$ if nessesary, we may also assume that $y=4$ mod $10$.

Update 07.10.2024: Dmitry Ezhov now found a solution $$ (23499130751021842,252697047241468990409432149765008357514,-1075346360334969622883) $$ to (2). A solution of this size would be infisible to find by direct search, but nice ascent method for the related equation $y^2+10xyz+x^3-x-2=0$ has been used, see comments.

The next-shortest open cubic equation in [1] is
$$ y^2+7xyz+3x^3-2=0. \quad\quad\quad (3) $$ Modulo $4$ and $7$ analysis shows that $x=7$ mod $14$. After replacing $y$ with $-y$ if nessesary, we may also assume that $y=3$ mod $14$. Does this equation have any integer solutions? Can a solution be found by first developing ascent formulas for the solutions to $y^2+xyz+3x^3-2=0$?

[1] Bogdan Grechuk, Polynomial Diophantine equations. A systematic approach, Springer, 2024