Find positive integer solutions for $n^4(n+1)+1=7^m$ This is the statement: 

Find all pairs of natural numbers $(n,m)$ that satisfies equation $$n^4(n+1)+1=7^m$$ 

I found one solution (2,2) and believe there is no more solutions. I'm beginner in number theory and don't know how to even start this. Can anyone help?

Let $d=\gcd(n^3-n+1,n^2+n+1)$.
Since $d\mid n^3-n+1$ and $d\mid n^2+n+1 \implies d\mid n^3-1$, so $$d\mid (n^3-1)-(n^3-n+1)= n-2$$ 
So $$d\mid (n^2+n+1) -(n-2)- (n^2-4)=7$$ So $d=1$ or $d=7$.
Case 1 $\boxed{d=1}$: 


*

*$n^2+n+1= 7^m$ and $n^3-n+1=1$ (so $n=1$ which does not work) or 

*$n^2+n+1= 1$ (impossible) and $n^3-n+1=7^m$
Case 2 $\boxed{d=7}$: 


*

*$n^2+n+1= 7^{m-1}$ and $n^3-n+1=7$ (so $n=2$ and $m=2$) or 

*$n^2+n+1= 7$ (so $n=2$) and $n^3-n+1=7^{m-1}$ (so $m=2$).
Conclusion. Only solution is $n=2$ and $m=2$.
