Let $P_n \ (n=3,4,5,6,7)$ be the set of positive integers $n^k+n^l+n^m$, where $k,l,m$ are positive integers. Find $n$ such that:

i) In the set $P_n$ there are infinitely many squares.

ii) In the set $P_n$ there are no squares.
1. **Analyzing \( n = 3 \):**
   - Consider \( P_3 \) which consists of numbers of the form \( 3^k + 3^l + 3^m \).
   - Setting \( k = l = m \), we get \( 3^k + 3^k + 3^k = 3 \cdot 3^k = 3^{k+1} \).
   - Since \( 3^{k+1} \) is not a square for any integer \( k \), we need to find another combination.
   - However, if we set \( k = l \) and \( m = k+1 \), we get \( 3^k + 3^k + 3^{k+1} = 2 \cdot 3^k + 3^{k+1} = 3^k (2 + 3) = 3^k \cdot 5 \), which is not a square.
   - Therefore, \( n = 3 \) does not produce infinitely many squares.

2. **Analyzing \( n = 4 \):**
   - Consider \( P_4 \) which consists of numbers of the form \( 4^k + 4^l + 4^m \).
   - Note that \( 4^k + 4^l + 4^m = 2^{2k} + 2^{2l} + 2^{2m} \).
   - Setting \( k = l \) and \( m = k+1 \), we get \( 4^k + 4^k + 4^{k+1} = 2 \cdot 4^k + 4^{k+1} = 2 \cdot 2^{2k} + 2^{2k+2} = 2^{2k+1} + 2^{2k+2} = 2^{2k+1} (1 + 2) = 2^{2k+1} \cdot 3 \), which is not a square.
   - Therefore, \( n = 4 \) does not produce infinitely many squares.

3. **Analyzing \( n = 5 \):**
   - Consider \( P_5 \) which consists of numbers of the form \( 5^k + 5^l + 5^m \).
   - Note that \( 5^k + 5^l + 5^m \equiv 3 \pmod{4} \) since \( 5 \equiv 1 \pmod{4} \).
   - A square modulo 4 can only be 0 or 1, hence \( 5^k + 5^l + 5^m \) can never be a square.
   - Therefore, \( n = 5 \) produces no squares.

4. **Analyzing \( n = 6 \):**
   - Consider \( P_6 \) which consists of numbers of the form \( 6^k + 6^l + 6^m \).
   - Note that \( 6^k + 6^l + 6^m \equiv 8 \pmod{10} \) since \( 6 \equiv 6 \pmod{10} \).
   - A square modulo 10 can only be 0, 1, 4, 5, 6, or 9, hence \( 6^k + 6^l + 6^m \) can never be a square.
   - Therefore, \( n = 6 \) produces no squares.

5. **Analyzing \( n = 7 \):**
   - Consider \( P_7 \) which consists of numbers of the form \( 7^k + 7^l + 7^m \).
   - Setting \( k = l \) and \( m = k+1 \), we get \( 7^k + 7^k + 7^{k+1} = 2 \cdot 7^k + 7^{k+1} = 7^k (2 + 7) = 7^k \cdot 9 = (7^{k/2} \cdot 3)^2 \), which is a square.
   - Therefore, \( n = 7 \) produces infinitely many squares.

Conclusion:
i) In the set \( P_n \) there are infinitely many squares for \( n = 7 \).
ii) In the set \( P_n \) there are no squares for \( n = 5 \) and \( n = 6 \).

The final answer is \( \boxed{ n \in \{7\} } \) for infinitely many squares and \( n \in \{5, 6\} \) for no squares.