$(r^2-q)(r^2+q)=p $-->$ r^2-q=1,r^2+q=p$ --> $2q=p-1 , 2r^2=p+1 $ if $3|r$ --> $3=r $--> $p=17$--> $q=8$ which is not prime if $r=3k+1 $or $3k+2$ --> $3|2r^2-2$ --> $3|p-1$ --> $3|q$ -->$ 3=q$ -->$ q=3,p=7,r=2$

Different approach: One of $p,q,r~$ should be $2$ $p=2\Rightarrow 2=(r^2+q)(r^2-q)\Rightarrow \begin{cases} r^2+q=2\\r^2-q=1\end{cases} \Rightarrow 2r^2=3$: Impossible $q=2\Rightarrow p=(r^2+2)(r^2-2)\Rightarrow \begin{cases} r^2+2=p\\r^2-2=1\end{cases}\Rightarrow p=5\Rightarrow r^2=3$: Impossible $r=2\Rightarrow 16-q^2=p \Rightarrow \begin{cases} 4+q=p\\4-q=1\end{cases}\Rightarrow p=7, q=3$ $\Longrightarrow (p,q,r)=(7,3,2)$

One liner: $p=(r^2-q)(r^2+q)\implies q=r^2-1=(r-1)(r+1)$. As $q$ is prime, $r=2$, $q=3$ and $p=7$.

By, difference, of squares, $p=(r^2-q)(r^2+q)$, now $r^2=q+1$, because $p,$ is prime. Now, if you square, a prime, it is always $1\pmod{3}$, hence, $q \equiv r^2-1\equiv 0\pmod{3}$, but since $q$, is prime, it follows, $q=3,$ hence, $r=2$, and $p=7.$