Example 1 (Austrian Mathematical Competition 1988) Find the sum of all divisors $d=2^{a} \cdot 3^{b} (a, b$ being natural numbers) of $N=19^{88}-1$.
$$
\begin{aligned}
N & =19^{88}-1=(20-1)^{88}-1=(4 \cdot 5-1)^{88}-1 \\
& =-C_{88}^{1} 4 \cdot 5+C_{88}^{2} 4^{2} \cdot 5^{2}-C_{88}^{3} 4^{3} \cdot 5^{3}+\cdots-C_{88}^{87} 4^{87} 5^{87}+C_{88}^{88} 4^{88} 5^{88} \\
& =-2^{5} \cdot 55+2^{6} M=2^{5}(-55+2 M),
\end{aligned}
$$

where $M$ is an integer.
Therefore, in the standard factorization of $N$, the highest power of 2 is 5.
On the other hand,
$$
\begin{aligned}
N & =(2 \cdot 9+1)^{88}-1 \\
& =C_{88}^{1} 2 \cdot 9+C_{88}^{2} 2^{2} \cdot 9^{2}+\cdots+C_{88}^{88} 2^{88} \cdot 9^{88} \\
& =3^{2} \cdot 2 \cdot 88+3^{3} \cdot T \\
& =3^{2}(2 \cdot 88+3 T),
\end{aligned}
$$

where $T$ is an integer.
Therefore, in the standard factorization of $N$, the highest power of 3 is 2, i.e.,
$$
N=2^{5} \cdot 3^{2} \cdot L,
$$

where $L$ does not contain the factors 2 and 3.
Thus, the sum of all divisors of $N$ of the form $d=2^{a} \cdot 3^{b}$ is
$$
\left(2+2^{2}+2^{3}+2^{4}+2^{5}\right)\left(3+3^{2}\right)=744
$$