\begin{align*}
\text{Given}
\quad b^2-4ac=p^2 \space\implies\space p^2+4ac=b^2\\
\quad\text{we have}\qquad A^2+B^2=C^2
\\
\text{which is a Pythagorean triple where}\\
A=m^2-k^2\quad B=2mk\quad C=m^2+k^2\\
\end{align*}
In this case,
$\quad p=m^2-k^2\\
4ac=B^2=(2mk)^2=4m^2k^2\implies ac=m^2k^2\\
b=m^2+k^2.$
In any Pythagorean triple, $B= 4n\space\implies \space ac\in\mathbb{N^2}.\quad$ The only way this can work is if both of $\quad ac\quad$ are either $1, 2,\space$ or equal odd primes, or squares themselves.
$\textbf{Non Pythagorean solutions:}$
Other solutions have no pattern that "jumps out" as seen in this sample of
$(p,4\times ac,b)\space$ triples.
$$(3 , 4x2 ,1 )\quad
(4 , 4x3 ,2 )\quad
(5 , 4x6 ,1 )\quad
(5 , 4x4 ,3 )\\
(6 , 4x8 ,2 )\quad
(6 , 4x5 ,4 )\quad
(7 , 4x12 ,1 )\quad
(7 , 4x10 ,3 )\\
(7 , 4x6 ,5 )\quad
(8 , 4x15 ,2 )\quad
(8 , 4x12 ,4 )\quad
(8 , 4x7 ,6 )\\
(9 , 4x20 ,1 )\quad
(9 , 4x18 ,3 )\quad
(9 , 4x14 ,5 )\quad
(9 , 4x8 ,7 )\\
(10 , 4x24 ,2 )\quad
(10 , 4x21 ,4 )\quad
(10 , 4x16 ,6 )\quad
(10 , 4x9 ,8 )\\
$$
