The answer is $2019$.
Clearly, the answer should not be an integer greater than $2018$. Hence we may remove the last $4$ terms on the left-hand side of the equation.

Next, we may assume that $x = a(6!) + b(5!) + c(4!) + d(3!) + e(2!) + f(1!)$, where $a, b, c, d, e, f$ are nonnegative integers and $a \le 6, b \le 5, c \le 4, d \le 3, e \le 2, f \le 1$.
Substituting into the original equation gives
$$1237a + 206b + 41c + 10d + 3e + f = 2019.$$
Since $206b + 41c + 10d + 3e + f \le 1231$, we have $788 \le 1237a \le 2019$. This gives $a = 1$. Then we have
$$206b + 41c + 10d + 3e + f = 782.$$
Similarly, we find that $b = 3, c = 4, d = e = f = 0$. Hence,
$$x = 1(6!) + 3(5!) + 4(4!) = 1176$$
is the only solution.