Diophantine equation:$x^5+x^4+1=p^y$ 
find all triplets $(x,y,p)$ satisfying $x^5+x^4+1=p^y$ where x, y are positive integers and p is a prime.

My attempt:I didn't know how to start. So I tried finding some triplets. Interestingly, $(1,1,3)$ satisfies the given equation, but I am not able to find any more.
Next, I tried factorising the equation i.e. $x^5+x^4+1=(x^2+x+1)(x^3-x+1)=p^y$
Now I am stuck. I tried considering the gcd of the common factors but it does not help.
Any ideas??
 A: For anybody like me who is unable to view the solution at Diophantus Era begins! that Math Lover's question comment linked to because they're not a member of Brilliant and doesn't want to join, here is a solution.
First, apart from $x = 1$, which leads to the solution of $(1, 1, 3)$ you've already found, then both of the factors on the left are greater than $1$ and, thus, must be positive powers of $p$. This gives
$$x^2 + x + 1 \equiv 0 \pmod{p} \tag{1}\label{eq1A}$$
$$x^3 - x + 1 \equiv 0 \pmod{p} \tag{2}\label{eq2A}$$
Next, \eqref{eq2A} minus \eqref{eq1A} gives
$$x^3 - x^2 - 2x \equiv 0 \pmod{p} \implies x(x + 1)(x - 2) \equiv 0 \pmod{p}  \tag{3}\label{eq3A}$$
This means $x \equiv 0 \pmod{p}$, $x \equiv -1 \pmod{p}$ or $x \equiv 2 \pmod{p}$. The first $2$ cases give in \eqref{eq1A} that $1 \equiv 0 \pmod{p} \implies p = 1$, which is not allowed. With the third case, \eqref{eq1A} gives $7 \equiv 0 \pmod{p} \implies p = 7$ is the only possibility.
Checking $x = 2$ itself shows it works to give on the left side $49 = 7^2$, so $(2, 2, 7)$ is another solution. Next, consider $x \gt 2$, so
$$x = 7z + 2 \tag{4}\label{eq4A}$$
for some integer $z \ge 1$. Substituting this into the $x^2 + x + 1$ factor, and letting it be equal to $7^m$ for some integer $m \ge 1$, gives
$$\begin{equation}\begin{aligned}
7^m & = (7z + 2)^2 + (7z + 2) + 1 \\
& = 49z^2 + 28z + 4 + 7z + 3 \\
& = 49z^2 + 35z + 7 \\
& = 7(7z^2 + 5z + 1)
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
Thus, since $z \ge 1 \implies 7z^2 + 5z + 1 \gt 1$, then $7 \mid 7z^2 + 5z + 1$, so
$$5z + 1 \equiv 0 \pmod{7} \tag{6}\label{eq6A}$$
Next, letting $x^3 - x + 1 = 7^n$ for some integer $n \ge 1$ and using \eqref{eq4A} gives
$$\begin{equation}\begin{aligned}
7^n & = (7z + 2)^3 - (7z + 2) + 1 \\
& = (7)^3z^3 + 3(7^2)(2)z^2 + 3(7)(4)z + 8 - 7z - 1 \\
& = (7)^3z^3 + 6(7^2)z^2 + (11)(7)z + 7 \\
& = 7((7)^2z^3 + 6(7)z^2 + (11)z + 1)
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Similar to before, this results in
$$11z + 1 \equiv 0 \pmod{7} \implies 4z + 1 \equiv 0 \pmod{7} \tag{8}\label{eq8A}$$
Next, \eqref{eq6A} minus \eqref{eq8A} gives
$$z \equiv 0 \pmod{7} \tag{8}\label{eq9A}$$
However, this contradicts both \eqref{eq6A} and \eqref{eq8A} since it results in $1 \equiv 0 \pmod{7}$, showing there is no such positive integer $z$.
In summary, the only $2$ triplets $(x, y, p)$ that satisfy the equation are $(1, 1, 3)$ and $(2, 2, 7)$.
A: Continuing your approach note that from the equality
$$
(x^3-x+1)(x^2+x+1)=p^y
$$
we obtain that $x^3-x+1=p^n$ and $x^2+x+1=p^m$ for some nonnegative integers $m$ and $n$.
For $x=1$ and $x=2$ we have solutions $(x,y,p)\in\{(1,1,3),(2,2,7)\}$.
Now note that for $x\geq 3$ we have
$$
x^3-x+1=(x^3-1)-(x-2)=(x-1)(x^2+x+1)-(x-2)>x^2+x+1,
$$
so $p^n>p^m$ or $n>m$. Hence, $p^m\mid p^n$, so $x^2+x+1\mid x^3-x+1$. Since
$$
x^3-x+1=(x-1)(x^2+x+1)-(x-2)
$$
we have $x^2+x+1\mid x-2$. However, for $x\geq 3$ we have $0<x-2<x^2+x+1$, so there are no solutions in case $x\geq 3$.
Therefore, all solutions to this equation are $(x,y,p)\in\{(1,1,3),(2,2,7)\}$.
