Answer: 110

Solution:

If $p \mid n^4 + 1,$ then $n^8 \equiv 1$ and $n^4 \equiv -1 \pmod p,$ so $n$ has order $8$ modulo $p$ and $8 \mid p - 1$ (and $p = 2$ fails since $n^4 + 1 \equiv 2 \pmod 4$). The smallest prime $p \equiv 1 \pmod 8$ is $17,$ and indeed $2^4 = 16 \equiv -1 \pmod{17}.$ Because the derivative $4n^3$ is not divisible by $17$ at such an $n,$ each root lifts to a root modulo $17^2 = 289,$ so $p = 17.$

The fourth roots of $-1$ modulo $17$ are $\pm 2$ and $\pm 8.$ To lift $n = 8,$ set $n = 8 + 17t:$ modulo $289,$ $$\begin{aligned} &n^4 + 1 \equiv 8^4 + 1 \\ &\quad {}+ 4 \cdot 8^3 \cdot 17t \\ &= 17(241 + 2048t), \end{aligned}$$ so we need $241 + 2048t \equiv 3 + 8t$ $\equiv 0 \pmod{17},$ giving $t \equiv 6$ and $n \equiv 8 + 102 = 110 \pmod{289}.$

The same computation lifts $2,$ $15,$ and $9$ to $155,$ $134,$ and $179$ respectively, so the least positive $m$ is $110.$ Indeed $110^4 + 1 = 146410001$ $= 289 \cdot 506609.$