Find all positive integers $m$ for which there exist three positive integers $a,b,c$ such that $abcm=1+a^2+b^2+c^2$. 
1. **Claim**: The only such \( m \) is \( m = 4 \), achieved by \((a, b, c) = (1, 1, 1)\).

2. **Proof**: Assume \( a, b, c \) are all positive integers such that \( abcm = 1 + a^2 + b^2 + c^2 \).

3. **Step 1**: Show that \( a, b, c \) are all odd.
   - If all of them are even, then the LHS is even but the RHS is odd, a contradiction.
   - If two of them are even, then the LHS is a multiple of \( 4 \) but \( a^2 + b^2 + c^2 + 1 \equiv 0 + 0 + 1 + 1 \equiv 2 \pmod{4} \), which is a contradiction since a multiple of \( 4 \) cannot be \( 2 \pmod{4} \).
   - If one of them is even, then the LHS is even and the RHS is odd again, so we have another contradiction.
   - Thus, \( a, b, c \) are all odd.

4. **Step 2**: Conclude that \( 4 \mid m \).
   - Since \( a, b, c \) are odd, \( a^2, b^2, c^2 \equiv 1 \pmod{4} \).
   - Therefore, \( a^2 + b^2 + c^2 + 1 \equiv 1 + 1 + 1 + 1 \equiv 0 \pmod{4} \).
   - Hence, \( 4 \mid abcm \implies 4 \mid m \).

5. **Step 3**: Assume for contradiction that there exists \( m \neq 4 \) and positive integers \( a, b, c \) such that the condition is satisfied.
   - Let \((A, B, C)\) be the triplet with \( A + B + C \) minimized for this particular \( m \).
   - Since \( 4 \mid m \), we must have \( m \geq 8 \).

6. **Step 4**: Assume WLOG that \( A \geq B \geq C \).
   - Note that \( m = \frac{1}{ABC} + \frac{A}{BC} + \frac{B}{CA} + \frac{C}{AB} \).
   - Since \( \frac{1}{ABC} \leq 1 \), \( B \leq A \leq CA \), and \( C \leq A \leq BA \), we have that \( \frac{1}{ABC}, \frac{B}{CA}, \frac{C}{AB} \) are all less than or equal to \( 1 \).
   - Therefore, \( \frac{A}{BC} \geq m - 3 \implies A \geq (m - 3)BC \).

7. **Step 5**: Rewrite the original equation as \( A^2 - (mBC)A + (B^2 + C^2 + 1) = 0 \).
   - By Vieta's formulas, \( (mBC - A, B, C) \) is also a solution for this particular \( m \).
   - We can verify that all entries are positive integers: \( mBC - A = \frac{B^2 + C^2 + 1}{A} \), so \( mBC - A > 0 \) (and clearly \( mBC - A \) is an integer).
   - Thus, \( (mBC - A, B, C) \) is indeed a positive integer solution for this particular \( m \).

8. **Step 6**: Since \( A + B + C \) is minimal, we must have \( (mBC - A) + B + C \geq A + B + C \implies mBC \geq 2A \).
   - Since \( A \geq (m - 3)BC \) as derived earlier, this means that \( mBC \geq 2A \geq (2m - 6)BC \implies m \geq 2m - 6 \implies 6 \geq m \).
   - But since \( m \geq 8 \), this is a contradiction!

9. **Conclusion**: Thus, \( m = 4 \) is indeed the only solution, as desired. 

\(\blacksquare\)

The final answer is \( \boxed{ m = 4 } \).