If $p=q$ then $p^{p-1} (3p - 2) = 19$, impossible, so $p\neq q$. Modulo $p$ we have $-2 \equiv 19 \pmod{q}$, so $p\mid 21$, forcing $p\in \{3,7\}$. Modulo $q$ we have $3p \equiv 19 \pmod{q}$. Thus for $p=3$ we have $q\mid 10$, forcing $q\in \{2,10\}$, while for $p=7$ we have $q\mid 2$, forcing $q = 2$. Checking these possibilities yields $(p,q) = (3,2)$ or $(p,q) = (7,2)$.

If $p=q,$then $p^{p-1}(3p-2)=19.$Namely $p\mid 19\implies p=19.$Then $p^{p-1}(3p-2)>19$ which is absurd.Thus $p\neq q.$By FLT, $p^q\equiv p\pmod{q},q^{p-1}\equiv 1\pmod{p}.$Hence $3p-19\equiv 0\pmod{q},21\equiv 0\pmod{p}$ Therefore $p=3,7$ Case1:$p=3$ $-10\equiv 0\pmod{q}\implies q=2,5.$Checking each case,we see that only $q=2$ case satisfies the condition. Case2:$p=7$ $2\equiv 0\pmod{q}\implies q=2$ which satisfies the condition. From case1,2,the answer is $\boxed{(p,q)=(3,2),(7,2)}\blacksquare$