The only solution of the Diophantine equation $(1) \; 2^{2n+1} + 2^n + 1 = x^k,$ where $n$, $x$ and $k>1$ are positive integers, is $(n,x,k) = (4,23,2)$. Proof: Let $p$ be the least prime divisor of $k$. Then $m=x^{k/p}$ is an odd integer, thus (1) can be expressed as $(2) \; 2^{2n+1} + 2^n + 1 = m^p.$ Case 1: $p=2$. If $n=1$, then $m^2=11$ by (2), implying $n \geq 2$. According to (2) $(3) \; 2^{n-2}(2^{n+1} + 1) = \textstyle{\frac{m - 1}{2} \cdot \frac{m + 1}{2}}$. The fact that $\textstyle{GCD(\frac{m - 1}{2},\frac{m + 1}{2})=1}$ implies $\textstyle{2^{n-2} \mid \frac{m \pm 1}{2}},$ hence $\textstyle{\frac{m \pm 1}{2}} = 2^{n-2}t$ for a positive integer $t$. Then by (3) $(4) \; 2^{n+1} + 1 = t(2^{n-2}t \mp 1),$ or $(5) \; 2^{n-2} = \textstyle{\frac{1 \pm t}{t^2 - 8}}$. Clearly $t$ is odd by (4). Moreover $\textstyle{\frac{1 \pm t}{t^2 - 8}} \geq 1$ by (5) since $n \geq 2$. But $\mid \textstyle{\frac{1 \pm t}{t^2 - 8}} \mid < 1$ when $t=1$ and $t \geq 5$, implying $t=3$, which inserted in (3) yield $2^{n-2}=-2 < 1$ or $2^{n-2}=4=2^2$. Hence $n-2 = 2$, i.e. $n = 4$ and $m = 2^{n-1}t - 1 = 2^3 \cdot 3 - 1 = 23 = x^{k/2}$, implying $k=1$ and $x=23$ since 23 is a prime. So the only solution of (1) when $p=2$ is $(n,x,k)=(4,23,2)$. Case 2: $p > 2$. According to (2) $(6) \; 2^n(2^{n+1} + 1) = (m - 1)\sum_{i=0}^{p-1} m^i,$ where $\sum_{i=0}^{p-1} m^i \equiv \sum_{i=0}^{p-1} 1^i = \sum_{i=0}^{p-1} 1 = p \equiv 1 \pmod{2}$ since both $m$ and $p$ are odd. Therefore $2^n \mid m - 1$ by (6). Thus $m - 1 = 2^ns$ for a positive integer $s$, which inserted in (6) yields $2^{n+1} + 1 = s \sum_{i=0}^{p-1} (2^ns + 1)^i > 1 + 2^{2n},$ hence $2^{1-n} < 1$, which is obviously not true for $n \geq 2$. So there is no solution of (1) when $p > 2$. q.e.d.

See IMO 2006, problem 4 )