I claim the only such $m$ is $m=4$, achieved by $(a, b, c) = (1, 1, 1)$. Firstly, I claim that $a, b, c$ are all odd. To prove this, assume otherwise. Then, either all of them are even, two of them are even, or one of them are even. 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 $a^2+b^2+c^2+1\equiv 0+0+1+1\equiv 2\pmod{4}$, as odd squares are $0\pmod{4}$ and even squares are $1\pmod{4}$, which is a 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, as desired. From this we can conclude that $4\mid m$ since $a^2 + b^2 + c^2 + 1 \equiv 1 + 1 + 1 + 1 \equiv 0\pmod{4}$, so $4\mid abcm\iff 4\mid m$ (as $a, b, c$ are odd). Now, assume for the sake of contradiction that there exists a $m\neq 4$ and positive integers $a, b, c$ such that the condition is satisfied. Then, let $(A, B, C)$ be the pair with $A+B+C$ is minimized for this particular $m$. Since $4\mid m$, we must have $m\geq 8$. Now, 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 $\frac{1}{ABC}, \frac{B}{CA}$, and $\frac{C}{AB}$ are all less than or equal to $1$, so we must have $\frac{A}{BC}\geq m-3\iff A \geq (m-3)BC.$ Now, rewrite the original equation as $A^2 - (mBC)A + (B^2+C^2+1) = 0$. Then, by Vieta, $(mBC-A, B, C)$ is also a solution for this particular $m$. We can verify that all entries are positive integers -- we have that from Vieta again or just from the problem statement $mBC-A = \frac{B^2+C^2+1}{A}$, so $mBC-A > 0$ (and clearly $mBC-A$ is an integer), so $(mBC-A, B, C)$ is indeed a positive integer solution for this particular $m$. Since $A+B+C$ is minimal, we must have that $(mBC-A) + B + C \geq A + B + C \iff mBC \geq 2A.$ Since $A\geq (m-3)BC$ as we derived in the above paragraph, this means that $$mBC \geq 2A\geq (2m-6)BC\iff m \geq 2m-6\iff 6 \geq m,$$but since $m \geq 8$, this is a contradiction! Thus, $m = 4$ is indeed the only solution, as desired. $\blacksquare$