Solution:

If the fourth number is $\geq 12$, then the last three numbers must sum to at least $12 + 2 \cdot 12 + 2^{2} \cdot 12 = 84 > 79$. This is impossible, so the fourth number must be less than $12$. Then the only way we can have the required divisibilities among the first four numbers is if they are $1, 2, 4, 8$. So the last two numbers now sum to $79 - 15 = 64$. If we call these numbers $8a, 8ab$ ($a, b > 1$) then we get $a(1 + b) = a + ab = 8$, which forces $a = 2, b = 3$. So the last two numbers are $16, 48$.