Are there integers $x,y,z$ such that $$ 1 + 3 x^3 + x y^2 + 6 y z^2 = 0 \,\, ? \quad\quad (1) $$

If the length of an equation is the sum of degrees of monomials plus sum of logarithms of the coefficients, then this is currently the shortest cubic equation for which it is open whether it has any integer solutions. The previous shortest open cubic equation was this one Can $9xy$ divide $1+x^2+x^3+y^2$? , and it turns out to be solvable in integers, with the smallest known solution having over $20$ digits for each variable.

On the other hand, an equation $1+4x^3+xy^2+2yz^2=0$ similar to (1) has no integer solutions. This has been proved here On the shortest open cubic equation by noticing that $x$ is odd, then rewriting as $x+4x^4+t^2+2tz^2=0$ where $t=xy$ and proving that this equation has no integer solutions with odd $x$. However, for (1), the same method lead to equation $$ x+3x^4+t^2+6tz^2=0 \quad \quad (2) $$ which is solvable in integers: take for example $x = -2359$, $t = -375074$, $z = 6430$. For this solution, $x$ is not a divisor of $t$, and it seems difficult to understand whether (2) has a solution such that $x$ is a divisor of $t$. Hence, a different method is required.

Modulo $12$ analysis shows that $x$ is $5$ modulo $12$, while $y$ is even but not divisible by $3$. The search returned no solutions for $|x|$ up to $400$ millions.