Solve Diophantine Equation Show that $ x^3 - 6 = 25y^2  + 35y $ doesn't have any non-zero integer solution.
What I have tried
$$ 25y^2 +  35y - x^3 + 6 = 0 $$
solving for y we get
$$ y = \frac{-35 \pm \sqrt { 625 + 100 x^3} } {50} $$
so  I need to show that
$$ 25 + 4x^3= z^2 $$
doesn't have a solution
or if it does I need to show that
the fraction is not equal to an integer
but I can't proceed.

Alternative solution
Suppose $x^3 = 25y^2 + 35y + 6 = (5y+6)(5y+1).$
If $(5y+6)$ and $(5y+1)$ have a common factor, then this common factor must divide $5$.  If the common factor is $> 1$, then it must be $5$.  But this would imply that $5$ is a factor of $x^3$, which would imply that $5 | x^3$ which would imply that $5|(5y+6)(5y+1)$.
This is impossible.  Therefore $(5y+6)$ and $(5y+1)$ are relatively prime.  Therefore $(5y+6)$ and $(5y+1)$ must each be a perfect cube.  This is impossible, because for any positive integer $n$, 
$[(n+1)^3 - n^3] = 3n^2 + 3n + 1 > 5.$
Note: It is therefore trivial that for $(5y+6)$ and $(5y+1)$ any perfect cubes, $(a^3)$ and $(b^3),$ where each of $(a^3)$ and $(b^3)$ may be positive or negative, you can't have $|(a^3) - (b^3)| = 5.$
