<s>[INST] 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?
 [/INST] 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. 
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