Let $(a, b) = d$, $a = sd$, $b = td$, $(s, t) = 1$, $t > 1$, then $std^2 = d^k(s^k + t^k)$. So $k \geq 2$ and $st = d^{k-2}(s^k + t^k)$. Since $(st, s^k + t^k) = 1$, we have $st = d^{k-2}$. Therefore, any prime factor of $st$ can be divided by $d$.
If there is a prime factor $p$ of $s$ or $t$ no less than $11$, then $p$ divides $d$. So $p^2$ divides $a$ or $p^2$ divides $b$, but $p^2 > 100$, which is a contradiction.
So the prime factor of $st$ may be $2$, $3$, $5$ or $7$.
If there are at least three prime factors of $st$ among $2$, $3$, $5$, $7$, then there is a prime factor of $s$ or $t$ no less than $5$. And $d > 2 \times 3 \times 5 = 30$, so that $a$ or $b \geq 5d > 100$, which is a contradiction. The prime factor set of $st$ cannot be $\{3, 7\}$, otherwise, $a$ or $b \geq 7 \times 3 \times 7 > 100$, which is a contradiction.
Similarly, the prime factor set of $st$ cannot be $\{5, 7\}$.
Therefore, the prime factor set of $st$ can only be $\{2\}$, $\{3\}$, $\{5\}$, $\{7\}$, $\{2, 3\}$, $\{2, 5\}$, $\{2, 7\}$ or $\{3, 5\}$.

i) If the prime factor set of $st$ is $\{3, 5\}$, then $d$ can only be $15$. Then, $s = 3$, $t = 5$. So there is one good pair $(a, b) = (45, 75)$.

ii) If the prime factor set of $st$ is $\{2, 7\}$, then $d$ can only be $14$. Then $s = 2$, $t = 7$ or $s = 4$, $t = 7$. So there are two good pairs $(a, b) = (28, 98)$ and $(56, 98)$.

iii) If the prime factor of $st$ is $\{2, 5\}$, then $d$ can only be $10$ or $20$.
For $d = 10$, then $s = 2$, $t = 5$; $s = 1$, $t = 10$; $s = 4$, $t = 5$; $s = 5$, $t = 8$.
For $d = 20$, then $s = 2$, $t = 5$; $s = 4$, $t = 5$.
There are six good pairs.

iv) If the prime factor of $st$ is $\{2, 3\}$, then $d$ can only be $6$, $12$, $18$, $24$ or $30$.
For $d = 6$, $s = 1$, $t = 6$; $s = 1$, $t = 12$; $s = 2$, $t = 3$; $s = 2$, $t = 9$; $s = 3$, $t = 4$; $s = 3$, $t = 8$; $s = 3$, $t = 16$; $s = 4$, $t = 9$; $s = 8$, $t = 9$; $s = 9$, $t = 16$.
For $d = 12$, $s = 1$, $t = 6$; $s = 2$, $t = 3$; $s = 3$, $t = 4$; $s = 3$, $t = 8$.
For $d = 18$, $s = 2$, $t = 3$; $s = 3$, $t = 4$.
For $d = 24$, $s = 2$, $t = 3$; $s = 3$, $t = 4$. $d = 30$, $s = 2$, $t = 3$.
There are $19$ good pairs.

v) If the prime factor set of $st$ is $\{7\}$, then $s = 1$, $t = 7$, $d$ can only be $7$ or $14$.
So, there are two good pairs.

vi) If the prime factor set of $st$ is $\{5\}$, then $s = 1$, $t = 5$, $d$ can only be $5$, $10$, $15$ or $20$. So, there are four good pairs.

vii) If the prime factor set of $st$ is $\{3\}$ then we have the following:
when $s = 1$, $t = 3$, $d$ can only be $3$, $6$, $\dots$, or $33$;
when $s = 1$, $t = 9$, $d$ can only be $3$, $6$ or $9$;
when $s = 1$, $t = 27$, $d$ can only be $3$.
There are $15$ good pairs.

viii) If the prime factor set of $st$ is $\{2\}$, then we have the following:
when $s = 1$, $t = 2$, $d$ can only be $2$, $4$, $\dots$, or $50$;
when $s = 1$, $t = 4$, then $d$ can only be $2$, $4$, $\dots$, or $24$;
when $s = 1$, $t = 8$, then $d$ can only be $2$, $4$, $\dots$, or $12$;
when $s = 1$, $t = 16$, then $d$ can only be $2$, $4$ or $6$;
when $s = 1$, $t = 32$, then $d$ can only be $2$.
There are $47$ good pairs.

Therefore, there are all together $1 + 2 + 6 + 19 + 2 + 4 + 15 + 47 = 96$ good pairs. $\boxed{96}$