6 Let $p$ be a given odd prime, find all positive integer solutions $x, y$ of $p^{x}-y^{p}=1$.

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6. From $p^{x}=y^{p}+1=(y+1)\left(y^{p-1}-y^{p-2}+\cdots-y+1\right)$, we know that $y+1=p^{n}$, where $n$ is an integer. Since $y>0$, it follows that $n>0$. Therefore,
$$\begin{aligned}
p^{x} & =\left(p^{n}-1\right)^{p}+1 \\
& =p^{n p}-p \cdot p^{n(p-1)}+\mathrm{C}_{p}^{2} p^{n(p-2)}-\cdots-\mathrm{C}_{p}^{p-2} p^{2 n}+p \cdot p^{n}
\end{aligned}$$

It is easy to see that, except for the last term, all terms on the right-hand side of the above equation are divisible by $p^{n+2}$ (note that since $p$ is a prime, all $\mathrm{C}_{p}^{i}$ for $i=1, \cdots, p-2$ are divisible by $p$), so $p^{n+1}$ is the highest power of $p$ on the right-hand side of (1), hence it must be that $x=n+1$. At this point, (1) becomes
$$p^{n p}-p \cdot p^{n(p-1)}+\mathrm{C}_{p}^{2} p^{n(p-2)}-\cdots-\mathrm{C}_{p}^{p-2} p^{2 n}=0$$

When $p=3$, (2) becomes $3^{3 n}-3 \cdot 3^{2 n}=0$, yielding $n=1$, so $x=y=2$. If $p \geqslant 5$, noting that $\mathrm{C}_{p}^{p-2}$ is not divisible by $p^{2}$, it is easy to see that all terms on the left-hand side of (2) except the last one are divisible by $p^{2 n+2}$, but the last term is not divisible by $p^{2 n+2}$, indicating that (2) cannot hold. Therefore, the only solution to this problem is when $p=3$, with $x=y=2$.