Question: Find every pair of $(n,p)$ in which $n$ is a positive integer and $p$ is an odd prime number so that the sum of every positive divisor of $p^{2^n-1}$ is a square number.
It can be seen that the sum of every positive divisor of $p^{2^n-1}$ is: $$S = \frac{p^{2^n}-1}{p-1} = (p+1)(p^2+1)...(p^{2^{n-1}}+1)$$ For every integer $i,j$ with $0 \leq i < j < n$, we have $$gcd((p^{2^{i}}+1), (p^{2^{j}}+1))=2$$ and $p^{2^{j}}+1$ can be divided by 2 but not 4 (with $j \geq 1$). Thus we can rewrite $p^{2^{i}}+1$ as $2a_i$ for every $i$ $(0 \leq i < n)$, with $gcd(a_i,a_j)=1$ for every $i \neq j$ and $a_j$ is an odd number with $1 \leq j<n$. Then $$S = 2^n a_0a_1...a_{n-1}$$ However, since $S$ is a square number so $a_j$ must also be a square number with $1 \leq j<n$, and $a_0$ must either be a square number or 2 times a square number. This implies that $p^{2^{j}}+1$ is 2 times a square number (with $j \geq 1$), and $p+1$ must either be a square number or 2 times a square number. Therefore, there exist a positive integer $c_j$ so that $p^{2^{j}}+1 = 2c_j^2$ with $j \geq 1$.
Now, if $p+1$ is a square number, then $$p+1=c^2 \iff p = (c-1)(c+1) \iff c=2, p=3 \implies n=1 $$ Else if $p+1$ is 2 times a square number, then $p+1=2c^2$. We also have $p^2+1 = 2c_1^2$.
Here I am stuck. I have found that $p=7, n=2$ is a solution for this case $(7+1=8=2 \times 2^2, 7^2+1=50 =2 \times 5^2$, but I couldn't find any more solutions. If $p=7,n=2$ is the only solution for this case, how can I prove that there aren't any more solutions?
From $$2(c_1+c)(c_1-c)=2c_1^2-2c^2=(p^2+1)-(p+1)=p\cdot(p-1)$$ we find $p$ must be the largest factor on the left, i.e., $$ c_1 + c = p = 2 (c_1-c)+1\implies c_1=3c-1\implies p^2<2c_1^2<18c^2=9p+9\implies p<10$$