Three. (25 points) Find all pairs of positive integers $(a, b)$ such that $a^{3}=49 \times 3^{b}+8$.

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Obviously, $49 \times 3^{6}+8$ is an odd number. Therefore, $a$ is odd.
Also, $a^{3}=49 \times 3^{b}+8 \geqslant 49 \times 3+8>5^{3}$, so $a>5$.
$$
\begin{array}{l}
\text { By } a^{3}=49 \times 3^{b}+8 \\
\Rightarrow a^{3}-8=49 \times 3^{b} \\
\Rightarrow(a-2)\left(a^{2}+2 a+4\right)=7^{2} \times 3^{b} .
\end{array}
$$

Let $\left(a-2, a^{2}+2 a+4\right)=d$. Then $d$ is odd.
Since $a^{2}+2 a+4=(a-2)(a+4)+12$
$\Rightarrow d \mid 12 \Rightarrow d=1$ or 3.
If $d=1$, then
$$
\left\{\begin{array}{l}
a-2=7^{2}, \\
a^{2}+2 a+4=3^{b}
\end{array}\right.
$$

or $\left\{\begin{array}{l}a-2=3^{b}, \\ a^{2}+2 a+4=7^{2},\end{array}\right.$
neither has positive integer solutions.
If $d=3$, then
$$
\left\{\begin{array}{l}
a-2=3 \times 7^{2}, \\
a^{2}+2 a+4=3^{b-1}
\end{array}\right.
$$

or
$$
\begin{array}{l}
\left\{\begin{array}{l}
a-2=3^{b-1}, \\
a^{2}+2 a+4=3 \times 7^{2}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a=11, \\
b=3 .
\end{array}\right.
\end{array}
$$

Therefore, there is only one pair of positive integers that satisfies the condition, which is $(11,3)$.