Because of the primality of $19$,$\newcommand{\f}{\mathbb{F}}$
a field $K$ is the splitting field for $f:=X^{19}-1$ over $\f_7$
if and only if $K$ contains a primitive $19$-th root of unity over $\f_7$.
Let $K$ be the splitting field of $f$ over $\f_7$,
and $\alpha$ be a primitive $19$-th root of unity.
We have $K=\f_{7^n}$ for some $n$ by the uniqueness of finite fields.
By Lagrange theorem,
the multiplicative order of $\alpha$ divides $|K^\times|=7^n-1$,
which translates to $19\mid 7^n-1$.
It follows that $n$ is the least positive integer solution to $7^n-1\equiv 0\ (\text{mod}\ 19)$.
An easy calculation shows that $n=3$.
Since the degree of any root of $f$ must divide $[K:\f_7]=3$,
this shows that the degree of any irreducible factor of $f$ is $1$ or $3$.
Also, $f$ splits completely over $\f_{7^3}$ because $\f_{7^3}$ is the splitting field for $f$.
By the way, the $\f_7$-factorization of $f$ is
$$
(X + 6) (X^{3} + 2 X + 6) (X^{3} + 3 X^{2} + 3 X + 6) (X^{3} + 4 X^{2} + X + 6) (X^{3} + 4 X^{2} + 4 X + 6) (X^{3} + 5 X^{2} + 6) (X^{3} + 6 X^{2} + 3 X + 6)
$$
and the $\f_{7^3}$-factorization of $f$ is
$$
(X + 6) (X + 2\alpha + 4) (X + 2\alpha + 6) (X + 6\alpha) (X + \alpha^2 + 5\alpha + 3) (X + \alpha^2 + 5\alpha + 5) (X + 2\alpha^2 + 4\alpha) (X + 2\alpha^2 + 6\alpha) (X + 3\alpha^2 + \alpha + 4) (X + 3\alpha^2 + 4\alpha + 6) (X + 3\alpha^2 + 5\alpha + 5) (X + 4\alpha^2 + 3) (X + 4\alpha^2 + 3\alpha + 2) (X + 5\alpha^2 + \alpha + 1) (X + 5\alpha^2 + 3\alpha + 1) (X + 5\alpha^2 + 6\alpha + 3) (X + 6\alpha^2) (X + 6\alpha^2 + 5) (X + 6\alpha^2 + 3\alpha + 2),
$$
where $\alpha^3+2\alpha+6=0$.
