Notice, that from conditions of the task, it follows, that $d_1 \cdot d_l = d_2 \cdot d_{l-1} = n$. Using this equality and first equality from conditions, it is possible to rewrite the second equality from the task:
$$N^b = d_{l-1} + d_l = \frac{n}{d_2} + \frac{n}{d_l} = \frac{n(d_1+d_2)}{d_1d_2} = \frac{nN^a}{d_1d_2} \text{ or } n = d_1d_2N^{b-a}.$$

Apart from that, if $b=a$, then $d_1+d_2 = N^a = N^b = d_{l-1} + d_l$, so $d_1 = d_{l-1}$; $d_2 = d_l$. Then the number needed has four divisors, or exactly two proper divisors. There are two forms of such numbers: $n = pq$, where $p, q$ are prime numbers, or $n = p^3$, where $p$ is prime. So, we can consider only case $b > a$.

Consider two cases.

1. Let $N$ be odd number. Then from equality $d_1 + d_2 = N^a$ we have, that sum $d_1 + d_2$ is odd number, then from these divisors there is exactly one even, another is odd. So it is obvious, that $d_1 = 2$, as it is the least even divisor. Then $d_2 = N^a - 2$ must be prime number. Let $d_2 = N^a - 2 = p$, then $N^a = p + 2$. So, $n = 2 \cdot p \cdot N^{b-a}$.

a) If $N$ is not prime number, then as $N \le p+2$, it has prime divisor $q \le p+2$. Obviously, $q \ne p$, and so $q < p$. But this means, that for $b > a$ number $n = 2 \cdot p \cdot N^{b-a}$ is divisible by $q < p = d_2$, and then number $p$ can not be $d_2$. We get contradiction.

b) If $N$ is prime number, then for $a > 1$ we get, that it follows from equality $N^a = p+2$, that $N < p+2$, then it is proper divisor of number $n$, less than $p$, that is impossible. So, $a=1$, i.e. numbers $p$ and $N = p+2$ is pair of twin primes. And then $\forall b \ge 1$ we have, that number $n = 2 \cdot p \cdot (p+2)^{b-1}$ is solution. Really, $d_{l-1} = 2(p+2)^{b-1}$, $d_l = p(p+2)^{b-1}$ and $d_{l-1} + d_l = 2(p+2)^{b-1} + p(p+2)^{b-1} = (p+2)^b = N^b$.

2. Let $N$ be even number. As $n = d_1d_2N^{a-b}$, then $n$ is also even, i.e. $d_1 = 2$.
Then as $d_1 = 2$, then $d_2 = 4$, as sum of two the least divisors is even. Then $N^a = 6$. Then $N = 6$, $a = 1$, thus $n = 2 \cdot 4 \cdot 6^{b-1}$. If $b > a = 1$, then number 3 is divisor of number $n$, then $d_2 = 3$ – contradiction. Thus $b = a = 1$.