Recall that the perfect powers are those integers $m^k$ with $k,m\in\{2,3,\ldots\}$. I don't consider $0$ or $1$ as a perfect power.

Y. Bugeaud, M. Mignotte and S. Siksek [Annals of Math., 2006] proved that the only perfect powers in the Fibonacci sequence are $F_6=2^3$ and $F_{12}=12^2$.

Let $p(n)$ be the partition function in number theory. By the Hardy-Ramanujan formula, $$\lim_{n\to\infty}\frac{\log p(n)}{\sqrt n}=\pi\sqrt{\frac23}.$$ So, $p(n)$ eventually grows faster than any polynomial but slower than exponential functions.

**QUESTION**: Can $p(n)$ be a perfect power?

In 2013 I conjectured that this question has a negative answer and verified this for $n\le 15000$.

**Remark**. I also conjecture that no Bell number (or Franel number $f_n=\sum_{k=0}^n\binom nk^3$, or Apery number $A_n=\sum_{k=0}^n\binom nk^2\binom{n+k}k^2$) is a perfect power. For the prime-counting function $\pi(x)$, I guess that $\pi(2^3)=4$ is the only perfect power in the increasing sequence $\pi(2^n)\ (n=1,2,3,\ldots)$.