For which integers $0 \le k \le 9$ do there exist positive integers $m$ and $n$ so that the number $3^m + 3^n + k$ is a perfect square?
1. We need to determine for which integers \(0 \le k \le 9\) there exist positive integers \(m\) and \(n\) such that \(3^m + 3^n + k\) is a perfect square.
2. We start by testing specific values of \(k\) to see if they work:
   - For \(k = 0\):
     \[
     3^3 + 3^2 + 0 = 27 + 9 = 36 = 6^2
     \]
     Thus, \(k = 0\) works.
   - For \(k = 3\):
     \[
     3^1 + 3^1 + 3 = 3 + 3 + 3 = 9 = 3^2
     \]
     Thus, \(k = 3\) works.
   - For \(k = 4\):
     \[
     3^2 + 3^1 + 4 = 9 + 3 + 4 = 16 = 4^2
     \]
     Thus, \(k = 4\) works.
   - For \(k = 6\):
     \[
     3^3 + 3^1 + 6 = 27 + 3 + 6 = 36 = 6^2
     \]
     Thus, \(k = 6\) works.
   - For \(k = 7\):
     \[
     3^2 + 3^2 + 7 = 9 + 9 + 7 = 25 = 5^2
     \]
     Thus, \(k = 7\) works.
3. Next, we check the values of \(k\) that do not work:
   - For \(k = 2, 5, 8\):
     - Consider the expression modulo 3:
       \[
       3^m + 3^n + k \equiv 0 + 0 + k \equiv k \pmod{3}
       \]
       Since a perfect square modulo 3 can only be 0 or 1, \(k\) must be 0 or 1 modulo 3. Therefore, \(k = 2, 5, 8\) do not work.
   - For \(k = 1\):
     - Consider the expression modulo 8:
       \[
       3^m + 3^n + 1 \pmod{8}
       \]
       Since \(3^m \equiv 3 \pmod{8}\) for \(m \geq 1\) and \(3^n \equiv 3 \pmod{8}\) for \(n \geq 1\), we have:
       \[
       3^m + 3^n + 1 \equiv 3 + 3 + 1 \equiv 7 \pmod{8}
       \]
       Since 7 is not a quadratic residue modulo 8, \(k = 1\) does not work.
   - For \(k = 9\):
     - If \(k = 9\) worked and \(m, n > 1\), we could divide the expression by 9 and get that \(k = 1\) works, which we have shown to be untrue. Therefore, \(k = 9\) does not work.
     - WLOG, let \(m = 1\), then we need to show \(3^n + 12\) cannot be a square. By modulo 4, we have to have \(n\) even, but then \(3^n\) and \(3^n + 12\) are both squares, which implies \(3^n = 4\), which is impossible.

Thus, the values of \(k\) that work are \(0, 3, 4, 6, 7\).

The final answer is \( \boxed{ k = 0, 3, 4, 6, 7 } \).