into the depressed cubic formula, we can obtain a solution by substituting
which turns the depressed cubic into a quadratic in $z^3$. Then the solution of the quadratic is
where $R$ is the discriminant of the quadratic, the "$b^2-4ac$" part of the problem.
The formula for the discriminant, $R$, is
For $a$, $b$ integers greater than one this is almost always negative, since it is negative if $a>2$ or $b>2$, so you have complex conjugate roots. If $a=b=2$ then the original equation is $x^3-8x-9=0$, which doesn't have any integer roots, since by the rational root theorem the roots must be $\pm1,3$, and these are not roots.
So you're looking for a formula involving the cube root of a square root of a negative number involving a series of squares, cubes and other powers, and hoping it will come out to be an integer.
At this point I decided to stop attempting this problem, since it looks like there is not much for it except to search for candidates using a computer, and I don't want to do that right now. However, I decided to post this partial answer since it might save someone some work. I'm sorry that I couldn't give a better answer, but I hope this at least moves things in the right direction.