Positive integer solutions of $x^2-1 = y^p$

Question

Let $x,y$ be two positive integers such that $x^2-1=y^p$ where $p$ is a prime. Find all possible $x,y,p$

I tried using Lifting the Exponent Lemma or considering the smallest prime that divides $x$ or that divides $y$ and tried solving. I think that $(x,y,p)=(3,2,3)$ is the only solution but there might be some more.

Instead of providing a full solution, may I get some hint at first? If I still can't solve, I'll post my methods I tried using the provided hint and then request for a full solution.

Answer 1

Here is a partial answer showing there are no solutions if $y$ is odd. I leave part of it in spoiler tags since the OP wanted hints.

We have $x^2 - 1 = y^p$ and so $(x + 1)(x - 1) = y^p$.
First assume that $y$ is odd.

We have $\gcd(x + 1, x - 1) \mid 2$, and since $y$ is odd, that means $x$ is even and thus $x \pm 1$ is odd. So they are coprime.

By the fact that $\mathbb{Z}$ is a UFD, this forces $x + 1 = m^p$ and $x - 1 = n^p$ for some positive integers $m, n$.

Thus $2 = m^p - n^p = (m - n)(m^{p-1}n + \ldots + m^2n^{p-2} + mn^{p-1})$.

This means $m = n + 1$ or $m = n + 2$.

In the first case we have $(n + 1)^p - n^p = 2$ which is impossible since the left side is odd.

In the second case we have $(n + 2)^p - n^p = 2$ which has unique solution $n = 1, p = 1$ but $1$ is not prime so this does not count. Note that the solution is unique because the LHS is an increasing function of both $n$ and $p$.

So there are no solutions with $y$ odd.