3.7 Find all prime numbers $p$ not exceeding 1000 such that $2p+1$ is a power of a natural number (i.e., there exist natural numbers $m$ and $n, \quad n \geqslant 2$, such that $2p+1=m^n$).
(Ukrainian Mathematical Olympiad, 1992)
［Solution］ If $2 p+1$ is a power of a natural number, i.e.,
$$
2 p+1=m^{n},
$$

then $m$ is an odd number.
Let $m=2 k+1$, then
thus
$$
\begin{array}{c}
2 p+1=(2 k+1)^{n} . \\
2 p=2 k \cdot A,
\end{array}
$$

where $p$ is a prime number, $A$ is an integer greater than 1,
i.e.,
$$
p=k \cdot A .
$$

Since $p$ is a prime number, $k=1$, hence $m=3$,
i.e.,
$$
2 p+1=3^{n} \text {. }
$$

Given $p \leqslant 1000$, then $3^{n} \leqslant 2001$.
And the powers of 3 not exceeding 2001 are
$$
3,3^{2}, 3^{3}, 3^{4}, 3^{5}, 3^{6} \text {, }
$$

Substituting these into equation (1) one by one, we find that only
$$
2 p+1=3^{3}
$$

makes $p$ a prime number, at this time $p=13$.
Thus, the only prime number that meets the requirements of the problem is $p=13$.