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?
 A: 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 $m=2$.
A: Use $$n^5+n^4+1=n^5-n^2+n^4-n+n^2+n+1=(n^2+n+1)(n^3-n+1).$$
Indeed, we obtain that there are non-negative integers $p$ and $q$, for which
$$n^2+n+1=7^p$$ and $$n^3-n+1=7^q.$$
Now, if $p=0$ or $q=0$ we obtain solutions $(0,0),$ $(-1,0)$ in integer numbers.
Let $p\geq1$ and $q\geq1$.
Thus, from the first equation we obtain $n=7k+2$ or $n=7k+4$ and the last is false for the second equation. 
Id est,
$$(7k+2)^2+7k+2+1=7^p$$ and
$$(7k+2)^3-(7k+2)+1=7^q$$ or
$$7k^2+5k+1=7^{p-1}$$ and $$49k^3+42k^2+11k+1=7^{q-1}.$$
Now, if $(p-1)(q-1)=0$, we obtain $k=0$ and the solution $(2,2).$
Let $p\geq2$ and $q\geq2$.
Thus, $11k+1-(5k+1)$ is divisible by $7$, which gives $k$ is divisible by $7$,
which is a contradiction because
$$7k^2+5k+1=7\left(k^2+\frac{5k}{7}\right)+1$$ is not divisible by $7$.
Thus, our equation has no another integer solutions. 
A: Here is a solution purely by modular arithmetic, given the factorization $(n^2+n+1)(n^3-n+1)$.
Both factors must be powers of $7$.  Suppose $n^3-n+1$ is a multiple of $49$.  That factor first has to be a multiple of $7$, and this is easily shown to require $n\equiv 2\bmod 7$.  Since the derivative $3n^2-1$ is nonzero $\bmod 7$, Hensel lifting applies and gives $n\equiv 37 \bmod 49$.  This fails to give a multiple of $49$ for the second factor $n^2+n+1$, and positive values of $n$ with this residue clearly give $n^2+n+1>7$.  No possibilities are left for $n^2+n+1$ to be a power of $7$.
So we can't have $n^3-n+1$ divisible by $49$ and make both factors equal to powers of $7$.  We are forced to accept $n^3-n+1\in\{1,7\}$, and the only positive values of $n$ satisfying this requirement are $n=1$ (which fails because $n^2+n+1$ is not a power of $7$) and $n=2$ (which is then the sole survivor).
