$b=a^p+1$ is a perfect square. Show that $p|(b-9)$

$p$ is a prime, and $a$ is a positive integer, $b=a^p+1$ is a perfect square. Show that $p|(b-9)$

It seem very interesting problem.if let $a^p+1=x^2$,it is clear $p\neq 2$

I have prove : $x$ is odd

proof:if $x$ is even number,then $(x+1,x-1)=1$,and note $a^p=(x+1)(x-1)$,then exist $r>s\ge 1\in N^{+}$ such $x+1=r^p,x-1=s^p$ so $$2=(x+1)-(x-1)=r^p-s^p=(r-s)(r^{p-1}+r^{p-2}s+\cdots+s^{p-1}) \ge p\ge 3$$which is contradiction

which is a pretty interesting and nice result. I wonder in which ways we may approach it.

From the comments we have the elementary proofs that $x^2 = a^p + 1$ has no solutions for prime $p > 3$, so it suffices to prove that when $p = 3$, $x$ is divisible by $3$.

We have $\gcd(a+1, a^2-a+1) \mid 3$, so we can assume that they are coprime and hope for a contradiction. If they are coprime, then both are squares, say $$\begin{align*}a+1 &= n^2 \\a^2-a+1 &= m^2 \,. \end{align*}$$ But $a^2 - a + 1 = m^2$ is impossible when $a > 1$, because $$(a-1)^2 = a^2 - 2a + 1 < a^2 - a - 1 < a^2 \,.$$ This is our contradiction.