$Case-1:$ $p=2$ gives $n=1$ $Case-2:$ $p>2$ $p(p-1)^2 = 3^n - 1$ $4 | RHS$ which means $n$ is even. $n=2m$ $(p-1)^3 + 1^3$ = $(3^m)^2 - (p-1)^2$ updating soon..

If $p=2$, then obviously $n=1$. From now on assume $p\not = 2$ Factoring we have: $p(p-1)^2 = 3^n - 1$. $p-1$ is odd, hence $4 \vert LHS \implies 4 \vert 3^n - 1 \implies n=2k$. Now $p(p-1)^2 = (3^k - 1)(3^k + 1)$. Obviously $p$ can divide only one of the factors on the right and also both of them are even. Now consider the two cases: Case 1: $p \vert 3^k - 1$ Now there exists integer $r \ge 2$, s.t $3^k - 1 = pr$. Substituting we have: $$p(p-1)^2 = pr(pr+2) \iff p^2 - (2+r^2)p + (1-2r) = 0$$ Consider this as a quadratic equation wrt p. Hence the discriminant has to be a square. But: $$D = r^4 + 4r^2 + 8r \implies (r^2 + 3)^2 > D > (r^2 + 2)^2$$ Hence D can never be a square. Case 2: $p \vert 3^k + 1$ Similarly there exists integer $r \ge 2$, s.t $3^k + 1 = pr$. Substituting we have: $$p(p-1)^2 = pr(pr -2) \iff p^2 - (2+r^2)p + (1+2r) = 0$$ Again the discriminant has to be a square. $D = r^4 + 4r^2 - 8r$ and since $r \ge 2$ we have: $(r^2 + 2)^2 > D \ge (r^2)^2$. So we have two options. $$D = (r^2 + 1)^2 \implies 2r^2 - 8r - 1 = 0 \implies r \not \in \mathbb{Z}$$ $$D = r^4 \implies 4r(r-2) = 0 \implies \boxed{r=2}$$ Now from this $3^k + 1 = 2p$, while $3^k - 1 = \frac{(p-1)^2}2$. Subtract them: $2 = 2p - \frac{(p-1)^2}2 \implies p=5, 1$. Obviously $p=5$ is the only acceptable value. Pluggin back in we get $n=4$. So finaly we have: $\boxed{ (p,n) = \{(2,1),(5,4)\}}$ as the only solutions. P.S. Special thanks to my friend Andrej