Answer: 97

Small Hint:

If $p$ divides $2019^8 + 1,$ then $2019^8 \equiv -1 \pmod{p},$ so the order of $2019$ modulo $p$ is exactly $16$

Big Hint:

The order divides $p - 1,$ so $p \equiv 1 \pmod{16};$ test the smallest such primes by repeated squaring of $2019 \bmod p$

Solution:

Suppose an odd prime $p$ divides $2019^8 + 1.$ Then $2019^8 \equiv -1 \pmod{p},$ so $2019^{16} \equiv 1$ while $2019^8 \not\equiv 1:$ the multiplicative order of $2019$ modulo $p$ is exactly $16.$ Since the order divides $p - 1,$ we need $p \equiv 1 \pmod{16},$ and the smallest such primes are $17$ and $97.$

Modulo $17:$ $2019 \equiv 13,$ and $13^2 = 169 \equiv -1,$ so $2019^8 \equiv (-1)^4 = 1$ and $2019^8 + 1 \equiv 2 \neq 0.$ Modulo $97:$ $2019 \equiv -18,$ and squaring repeatedly, $$\begin{aligned} 2019^2 &\equiv 324 \equiv 33, \\ 2019^4 &\equiv 33^2 = 1089 \\ &\equiv 22, \\ 2019^8 &\equiv 22^2 = 484 \\ &\equiv -1 \pmod{97}. \end{aligned}$$

So $97$ divides $2019^8 + 1,$ and it is the least odd prime factor: $97.$