, 2012

If $(2p)^n + 1 = a^3$, then we have
$$(2p)^n = a^3 - 1 = (a-1)(a^2+a+1) = (a-1)[(a-1)^2+3(a-1)+3].$$
Since $a$ is odd, it follows that $a^2+a+1$ is also odd, hence $2^n \nmid a-1$. This means that we have $a = 2^n p^k + 1$ for some integer $k \ge 0$.

We obtain
$$2^n p^n = 2^n p^k (2^{2n} p^{2k} + 3 \cdot 2^n p^k + 3). \quad (1)$$

Case 1. $k > 0$. Clearly, we have $k < n$, hence from (1) it follows $p|2^{2n} p^{2k} + 3 \cdot 2^n p^k + 3$. We get $p|3$, not possible since $p \ge 5$.

Case 2. $k = 0$. From (1) we obtain
$$p^n = 4^n + 3 \cdot 2^n + 3. \quad (2)$$
For $n = 1$ we have $p = 13$.

For $n = 2$ we have $p^2 = 31$, not possible.

For $n = 3$ we have $p^3 = 64 + 24 + 3 < 5^3$, not possible.

For $n \ge 3$ we prove by induction that
$$5^n > 4^n + 3 \cdot 2^n + 3 = p^n,$$
and so $p < 5$, which is not possible.

The unique solutions are $n = 1, p = 13$.