Nice problem. Factorising, $(p-q)(p+q)=r^n$. Let $p-q=r^a$ and $p+q=r^b$, where $a+b=n$ and obviously $a<b$. Thus $2p=r^{a}(r^{b-a}+1)$. Case 1) If $a\geq 1$, then $r=2$ or $r=p$, since $r\mid 2p$. Subcase 1) If $r=p$, then if $a=1$ and we get that $2=r^{b-1}+1\implies b=1$, hence $n=2$: $p-q=r$ and $p+q=r$, thus $2q=0\implies q=0$, contradiction. If $a=2$, then $r=p=2$ and this is not true since $r^{b-2}\neq 0$. If $a\geq 3$, then obviously, $p$ is not a prime anymore and contradiction. Subcase 2) If $r=2$, then $a=0,1,2$, otherwise $p$ is not a prime anymore and contradiction. If $a=2$, then $p=2$, but this gives a contradiction since $r^{b-2}\neq 0$. If $a=1$, then $p=2^{n-2}+1$ and $p-q=2$ and $p+q=2^{n-1}$, $q=2^{n-2}-1$. If $n$ is even, then $q=2^{n-2}-1\equiv (-1)^{2k}-1\equiv 0\pmod 3\implies q=3$, thus $p=5$ and hence $n=4$. If $n$ is odd, then $p=2^{n-2}+1\equiv (-1)^{2k+1}+1\equiv 0\pmod 3\implies p=3$, thus $q=1$, contradiction. If $a=0$, then contradiction by modulo 2 on $2p=2^{n}+1$. Case 2) If $a=0$, then $p=q+1$ and only consecutive primes are $2$ and $3$, thus $q=2$, $p=3$ and we have that $r^n=5$, thus $n=1$ and $r=5$. Answer. $\boxed{(p, q, r, n)=(3,2,5,1)\text{ and } (p, q, r, n)=(5,3,2,4)}$.