An integer number $m\geq 1$ is [i]mexica[/i] if it's of the form $n^{d(n)}$, where $n$ is a positive integer and $d(n)$ is the number of positive integers which divide $n$. Find all mexica numbers less than $2019$.

Note. The divisors of $n$ include $1$ and $n$; for example, $d(12)=6$, since $1, 2, 3, 4, 6, 12$ are all the positive divisors of $12$.

[i]Proposed by Cuauhtémoc Gómez[/i]
To find all mexica numbers less than \(2019\), we need to identify numbers of the form \(n^{d(n)}\), where \(d(n)\) is the number of positive divisors of \(n\). We will consider different values of \(d(n)\) and find the corresponding \(n\) such that \(n^{d(n)} < 2019\).

1. **Case \(d(n) = 1\):**
   - The only number with exactly one divisor is \(n = 1\).
   - Therefore, \(1^{d(1)} = 1^1 = 1\).

2. **Case \(d(n) = 2\):**
   - Numbers with exactly two divisors are prime numbers.
   - Let \(p\) be a prime number. Then \(p^{d(p)} = p^2\).
   - We need \(p^2 < 2019\). The largest prime \(p\) such that \(p^2 < 2019\) is \(43\) (since \(43^2 = 1849\) and \(47^2 = 2209\)).
   - The primes less than or equal to \(43\) are: \(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43\).

3. **Case \(d(n) = 3\):**
   - Numbers with exactly three divisors are squares of prime numbers.
   - Let \(p\) be a prime number. Then \(n = p^2\) and \(d(n) = 3\).
   - We need \((p^2)^3 = p^6 < 2019\). The largest prime \(p\) such that \(p^6 < 2019\) is \(2\) (since \(2^6 = 64\) and \(3^6 = 729\)).
   - Therefore, \(2^6 = 64\).

4. **Case \(d(n) = 4\):**
   - Numbers with exactly four divisors can be of the form \(p^3\) or \(pq\) where \(p\) and \(q\) are distinct primes.
   - For \(p^3\):
     - We need \(p^{3 \cdot 4} = p^{12} < 2019\). The largest prime \(p\) such that \(p^{12} < 2019\) is \(2\) (since \(2^{12} = 4096\) which is greater than 2019).
     - Therefore, there are no such numbers.
   - For \(pq\):
     - We need \((pq)^4 < 2019\). The largest product \(pq\) such that \((pq)^4 < 2019\) is \(6\) (since \(6^4 = 1296\) and \(7^4 = 2401\)).
     - Therefore, \(6^4 = 1296\).

Combining all the cases, the mexica numbers less than \(2019\) are:
\[1, 2^2, 3^2, 5^2, 7^2, 11^2, 13^2, 17^2, 19^2, 23^2, 29^2, 31^2, 37^2, 41^2, 43^2, 2^6, 6^4\]
\[= 1, 4, 9, 25, 49, 121, 169, 289, 361, 529, 841, 961, 1369, 1681, 1849, 64, 1296\]

The final answer is \( \boxed{ \{1, 4, 9, 25, 49, 121, 169, 289, 361, 529, 841, 961, 1369, 1681, 1849, 64, 1296\} } \)