Suppose that the positive divisors of a positive integer $n$ are $1=d_1<d_2<\ldots<d_k=n$, where $k \geq 5$. Given that $k \leq 1000$ and $n={d_2}^{d_3}{d_4}^{d_5}$, compute, with proof, all possible values of $k$.
1. **Claim 1: There are fewer than two primes that divide \( n \).**

   **Proof:**
   Suppose \( n \) has three distinct prime divisors \( p, q, r \) with \( p < q < r \). Then, the divisors of \( n \) include \( p, q, r \), and their products. Specifically, \( d_2 = p \), \( d_3 = q \), and \( d_4 \) must be at least \( qr \). However, since \( d_4 \geq qr \), \( q \) and \( r \) must be among the first few divisors, leading to a contradiction because \( d_3 \) should be less than \( d_4 \). Therefore, \( n \) cannot have three distinct prime divisors.

2. **Case 1: \( n \) has only one prime divisor \( p \).**

   If \( n \) has only one prime divisor \( p \), then the divisors of \( n \) are \( 1, p, p^2, p^3, \ldots, p^m \). Given \( k \geq 5 \), we have:
   \[
   d_2 = p, \quad d_3 = p^2, \quad d_4 = p^3, \quad d_5 = p^4
   \]
   Therefore,
   \[
   n = p^{p^2} \cdot p^{3p^4} = p^{p^2 + 3p^4}
   \]
   The number of divisors \( k \) of \( n \) is:
   \[
   k = p^2 + 3p^4 + 1
   \]
   Since \( k \leq 1000 \), we need to find \( p \) such that \( p^2 + 3p^4 + 1 \leq 1000 \).

   - For \( p = 2 \):
     \[
     k = 2^2 + 3 \cdot 2^4 + 1 = 4 + 48 + 1 = 53
     \]
   - For \( p = 3 \):
     \[
     k = 3^2 + 3 \cdot 3^4 + 1 = 9 + 243 + 1 = 253
     \]

   Therefore, the possible values of \( k \) in this case are \( 53 \) and \( 253 \).

3. **Case 2: \( n \) has two prime divisors \( p \) and \( q \) with \( p < q \).**

   If \( n \) has two prime divisors \( p \) and \( q \), then the divisors of \( n \) include \( 1, p, q, p^2, pq, q^2, \ldots \). Given \( k \geq 5 \), we have:
   \[
   d_2 = p, \quad d_3 = p^2, \quad d_4 = q, \quad d_5 = p^3
   \]
   Therefore,
   \[
   n = p^{p^2} \cdot q^{p^3}
   \]
   The number of divisors \( k \) of \( n \) is:
   \[
   k = (p^2 + 1)(p^3 + 1)
   \]
   Since \( k \leq 1000 \), we need to find \( p \) and \( q \) such that \( (p^2 + 1)(p^3 + 1) \leq 1000 \).

   - For \( p = 2 \):
     \[
     k = (2^2 + 1)(2^3 + 1) = 5 \cdot 9 = 45
     \]
   - For \( p = 3 \):
     \[
     k = (3^2 + 1)(3^3 + 1) = 10 \cdot 28 = 280
     \]

   Therefore, the possible values of \( k \) in this case are \( 45 \) and \( 280 \).

Combining both cases, the possible values of \( k \) are \( 45, 53, 253, \) and \( 280 \).

The final answer is \( \boxed{45, 53, 253, 280} \).