Senderov V.A.

Find all primes $p$ for each of which there exist natural numbers $x$ and $y$ such that $p^{x}=y^{3}+1$.
$y^{3}+1=(y+1)\left(y^{2}-y+1\right)$, so $y+1=p^{\alpha}, y^{2}-y+1=p^{\beta}$, where $\alpha>0, \beta \geq 0$.

First method. If $\beta=0$, then $y=1, p=2(x=1)$.

Let $\beta>0$. Then $p$ is a common divisor of the numbers $y+1$ and $y^{2}-y+1$, and therefore, of the numbers $y+1$ and $(y+1)^{2}-\left(y^{2}-y+1\right)$ $=3 y$.

Since $\gcd(y, y+1)=1, p=3$. This happens: $3^{2}=2^{3}+1$.

Second method. $p^{x}=\left(p^{\alpha}-1\right)^{3}+1=p^{\alpha}(3+A)$, where $A$ is divisible by $p$.

Since $3+A=p^{\gamma}(\gamma \geq 0)$, either $p=3$, or $\gamma=0$. In the latter case $p^{x}=p, x=1$. Thus, $y=y^{3}$, from which $y=1, p=2$.