13 If a positive integer $n$ can be written in the form $a^{b}$ (where $a, b \in \mathbf{N}, a \geqslant 2, b \geqslant 2$), then $n$ is called a "good number". Among the positive integers adjacent to the positive integer powers of 2, find all the "good numbers".
13 Let the required "good number" be $n$, which can be expressed as $k^{l}$ (where $k, l \in \mathbf{N}, k \geqslant 2, l \geqslant 2$). Since $n$ is adjacent to a positive integer power of 2, there exists a positive integer $t$ such that $2^{t} = k^{l} \pm 1$. Clearly, $k$ is an odd number.
(1) If $l$ is odd, then
$$
2^{t}=(k \pm 1)\left(k^{l-1} \mp k^{l-2}+k^{l-3}+\cdots \mp k+1\right),
$$

thus
$$
k^{l-1} \mp k^{l-2}+k^{l-3}+\cdots \mp k+1 \mid 2^{t}.
$$

Since $k$ and $l$ are both odd, the sum or difference of an odd number of odd numbers is always odd, so
$$
k^{l-1} \mp k^{l-2}+k^{l-3}+\cdots \mp k+1
$$

is also odd, and thus it can only be
$$
k^{l-1} \mp k^{l-2}+k^{l-3}+\cdots \mp k+1=1,
$$

substituting into (1) gives
$$
2^{t}=k^{l} \pm 1=k \pm 1,
$$

hence $l=1$, which contradicts $l \geqslant 2$.
(2) If $l$ is even, then $k^{l} \equiv 1(\bmod 4)$. If $2^{t}=k^{l}+1$, then
$$
2^{t}=k^{l}+1 \equiv 2(\bmod 4),
$$

thus $t=1$, so $k^{l}=2^{1}-1=1$, which is not a "good number". If
$$
2^{t}=k^{l}-1=\left(k^{\frac{l}{2}}+1\right)\left(k^{\frac{l}{2}}-1\right),
$$

then
$$
k^{\frac{l}{2}}+1\left|2^{t}, k^{\frac{l}{2}}-1\right| 2^{t}.
$$

Let $k^{\frac{l}{2}}+1=2^{p}$ and $k^{\frac{l}{2}}-1=2^{q}$, where $p, q \in \mathbf{N}, p>q, p+q=t$. Therefore,
$$
2=2^{p}-2^{q}=2^{q}\left(2^{p-q}-1\right),
$$

since $2^{p-q}-1$ is odd, so $2^{q}=2,2^{p-q}-1=1$, solving gives $p=2, q=1$, thus
$$
k^{l}=2^{t}+1=2^{p+q}+1=2^{3}+1=9.
$$

In summary, there is only one "good number", which is 9.