I don't think you got the second factor right.  Let $a=1989^{2^t}$, then the first factor is $a-1 = 1989^{2^t}-1$, and the second factor is:

$$a^{n-1} + a^{n-2} + \cdots + a + 1 = 1989^{2^t(n-1)} + 1989^{2^t(n-2)} + \cdots + 1989^{2^t} + 1.$$

When $n$ is an odd number, how many factor of $2$ do you think the second factor has?