Difficult diophantine equation Solve for integers:
$4n^4+7n^2+3n+6=m^3.$
Hi this is a problem from an Bulgarian olympiad for which I have no idea how to solve. 
I figured out using wolfram alpha that $16\cdot m^3-47$ must be a square number.
I would appreciate any solutions. Thank you in advance! 

The idea is going modulo $9$.
Indeed, we will prove that $4n^4 + 7n^2+3n+6$ leaves only remainders $2,5,6$ modulo $9$. None of these are cubes modulo $9$(only $0,1,8$ are), completing the proof that no such integers $n,m$ exist.
For this, we note that if $n \equiv 0 \pmod{3}$ then  $4n^4 + 7n^2+3n+6 \equiv 6\pmod{9}$.
If $n \equiv 1 \pmod{3}$  then $4n^4 + 7n^2+3n+6 \equiv 2 \pmod{9}$.
Finally, if $n \equiv - 1 \pmod{3}$ then $4n^4+7n^2+3n+6 \equiv 5 \pmod{9}$.
