Smallest $m>1$ such that the number of Collatz steps needed for $238!+m$ to reach $1$ differs from that for $238!+1$.

Question

Let $h(x)$ be the number of steps^ needed for $x$ to reach $1$ in the Collatz/3n+1 problem. I found that

$$h(238!+n)=h(238!+1), \;\; \forall 1 < n \leq 690,000,000$$ Here "!" is the standard factorial.

This is a lot of consecutive terms with the same height and beats the current record by far. Now I am wondering:

What is the smallest $m > 1$, such that $h(238!+m) \neq h(238!+1)$?

I don't know of an efficient way of finding it. We know that $h(2\cdot(238!+1))=h(238!+1)+1$, so $m \leq 238!+1$, but that's a rather large upper bound.

UPDATE 16/08/2021: Martin Ehrenstein found that $10^9 < m < 10^{94}$. See A346775. Later user mjqxxxx improved the upper bound to $m \leq 2^{64}$.

UPDATE 21/08/2021: Martin Ehrenstein improved the upper bound to $m < 11442739136455298475$.

^ $3x+1$ is considered one step and $x/2$ is one step.