find all p prime number such that : sum of divisor $p^{4}$ a perfect square

Question

Problem :

determine all prime numbers such that :

$1+p+p^{2}+p^{3}+p^{4}=m^{2}$ , $m\in \mathbb N$ My try :

$p=2$ so we find : $1+2+4+8+16=31≠m^{2}$

$p=3$ so we find :

$1+3+9+27+81=9^{2}$

$p=5$ so we find :

$1+5+25+125+625≠m^{2}$ $.$ $.$ $.$ so how many numbers ??

if we find : $m^{2}≤1+p+p^{2}+p^{3}+p^{4}≤(m+1)^{2}$

then we can say no prime number $p>3$

but how I find $m,m+1$ ??