3. Find the positive integer solutions $(p, n, k)$ of the equation $3^{p}+4^{p}=n^{k}$, where $p$ is a prime number, $k>1$.
3. Clearly, $3^{2}+4^{2}=5^{2}$, i.e., $p=2, n=5, k=2$ is a solution to the equation.
Assume henceforth that $p$ is an odd prime, $p=2 l+1$. Then
$$
\begin{aligned}
n^{k} & =3^{2 l+1}+4^{2 l+1} \\
& =(3+4)\left(3^{2 l}-3^{2 l-1} \times 4+3^{2 l-2} \times 4^{2}-\cdots+4^{2 l}\right) .
\end{aligned}
$$

Thus, $7\left|n^{k}, 7\right| n$.
Since $k>1$, we have $49 \mid n^{k}$, i.e.,
$$
3^{2 l+1}+4^{2 l+1} \equiv 0(\bmod 49) \text {. }
$$

By the binomial theorem,
$$
\begin{array}{l}
3^{2 l+1}=3 \times 9^{l}=3(7+2)^{l} \\
\equiv 3\left(l \times 7 \times 2^{l-1}+2^{l}\right) \\
\equiv(21 l+6) 2^{l-1}(\bmod 49), \\
4^{2 l+1}=4(14+2)^{l} \\
\equiv 4\left(l \times 14 \times 2^{l-1}+2^{l}\right) \\
\equiv(56 l+8) 2^{l-1}(\bmod 49) .
\end{array}
$$

Thus, $3^{2 l+1}+4^{2 l+1} \equiv(77 l+14) 2^{l-1}(\bmod 49)$.
Since $49 \mid \left(3^{2 l+1}+4^{2 l+1}\right)$, we have
$$
\begin{array}{l}
49|(77 l+14) \Leftrightarrow 7|(11 l+2) \\
\Leftrightarrow 7 \mid(4 l+2),
\end{array}
$$

i.e., $4 l+2 \equiv 0(\bmod 7)$.
The solution to this congruence is $l \equiv 3(\bmod 7)$.
Thus, $p=2 l+1 \equiv 0(\bmod 7)$.
Since $p$ is a prime, $p$ can only be 7.
Notice that,
$$
\begin{array}{l}
3^{7}+4^{7}=2187+16384 \\
=18571=49 \times 379 .
\end{array}
$$

But 379 is a prime, so the above expression cannot be written in the form $n^{k}$ $(k \geqslant 2)$, i.e., there is no solution when $p$ is an odd prime.
In summary, the equation has only one positive integer solution
$$
(p, n, k)=(2,5,2) \text {. }
$$