Three. (25 points) Find all integer pairs $(a, b)$ that satisfy $a^{3}=7 \times 3^{b}-64$.

---

Please note that the translation retains the original formatting and structure of the text.
Obviously, $b$ cannot be a negative integer.
When $b=0$, $a^{3}=-57$, at this time, $a$ is not an integer, which does not meet the requirements of the problem.

When $b>0$, obviously, $7 \times 3^{b}-64$ is an odd number, thus, $a$ is an odd number.
$$
\begin{array}{l}
\text { By } a^{3}=7 \times 3^{b}-64 \\
\Rightarrow a^{3}+64=7 \times 3^{b} \\
\Rightarrow(a+4)\left(a^{2}-4 a+16\right)=7 \times 3^{b} .
\end{array}
$$

Note that, $a+4$ and $a^{2}-4 a+16$ are both odd numbers,
$$
a^{2}-4 a+16=(a-2)^{2}+12>0 \text {. }
$$

Thus, $a+4>0$.
$$
\text { Let }\left(a+4, a^{2}-4 a+16\right)=d \text {. }
$$

Then $d$ is an odd number.
$$
\begin{array}{l}
\text { By } a^{2}-4 a+16=(a+4)(a-8)+48 \\
\Rightarrow d \mid 48 \Rightarrow d=1,3 .
\end{array}
$$

When $d=1$,
$$
\begin{array}{l}
\left(a+4, a^{2}-4 a+16\right) \\
=\left(1,7 \times 3^{b}\right),\left(7 \times 3^{b}, 1\right),\left(7,3^{b}\right),\left(3^{b}, 7\right) .
\end{array}
$$

Upon inspection, none of these have integer solutions, which does not meet the requirements of the problem.
When $d=3$,
$$
\begin{array}{l}
\left(a+4, a^{2}-4 a+16\right) \\
=\left(3,7 \times 3^{b-1}\right),\left(7 \times 3^{b-1}, 3\right), \\
\quad\left(3 \times 7,3^{b-1}\right),\left(3^{b-1}, 3 \times 7\right) .
\end{array}
$$

Upon inspection, only $(a, b)=(-1,2),(5,3)$ meet the requirements of the problem.

In summary, all the integer pairs $(a, b)$ that meet the conditions are $(-1,2)$ or $(5,3)$.