$b=a^p+1$ 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.

I am in 8th grade, so please if I make mistakes in the calculation then please tell me, I would delete the answer.
You proved that $x$ is odd. That would mean that $x^2$ is odd too.
So the equation $x^2=a^p+1$
If we subtract $1$ from both sides then $a^p$ would be even. Which implies that $a$ is even.
Since $x$ is odd, $x=2k+1$ and since $a$ is even, $a=2n$
Putting these in the equation, we get:
$(2k+1)^2=(2n)^p+1$
$\implies 4k^2+4k+1=2^p n^p+1$
$\implies k(k+1)=2^{p-2}n^p$
Since one of $k$ and $k+1$ has to be even
So assume $k$ is even, then $k+1$ is odd.
Now consider the following:
If $n^p$ is even
Then $n$ is even
$2l(2l+1)=2^{p-2}n^p$
$\implies l=2m$
$\implies k=4m$
$\implies k\geq 4$
$\implies x\geq 9$
$\implies b-9 \geq 72$-----------$(1)$
AND, If $n^p$ is odd
Then $n$ is odd, so let $n=2\alpha+1$
By the comment of John L and Kieren MacNillan, it seems that $p$ cannot be greater than $3$ so consider the case $p=3$:
$2l(2l+1)=2(2\alpha+1)^3$
$\implies 2l^2+l=8\alpha^3+6\alpha^2+6\alpha+1$
$\implies l-1=2\beta$
$\implies l\geq 3$
$\implies k\geq 6$
$\implies x\geq 13$
$\implies b-9\geq 160$---------$(2)$
Assuming that the statement in the question is true, 3 cannot divide 160. Which means that $n$ can't be even.
So $n$ is odd.
So from $(1)$, we get $b\geq 81$
$\implies (2n)^3+1\geq 81$
$\implies n\geq\sqrt[3]{10}$
Since $n$ is a positive integer,
$n\geq\lceil \sqrt[3]{10} \rceil=3$
$\implies a\geq 6$
If the statement is not correct then $(2)$ can also be the case, then from $(2)$ we get $b\geq 169$
$\implies (2n)^3+1\geq 169$
$\implies n\geq\sqrt[3]{21}$
Since $n$ is a positive integer,
$n\geq\lceil\sqrt[3]{21}\rceil=3$
$\implies a\geq 6$
So in both cases $a\geq 6$
These followings I derived in this answer might help you prove the statement:
$●b-9\geq 72$ if the statement is correct. Otherwise $b-9\geq 72$ and $b-9\geq 160$ both could be the case.
$●a=2n\geq 6$ where $n$ is an odd number.
$●$The only $p$ to consider is $p=3$ by the comment of John L and Kieren MacMillan.
