Let us write $N(k)$ for the number of consecutive composite numbers following $k! + 1$.
If $m$ is the smallest prime greater than $k$, then for all $n$ with $1 < n < m$, $n$ must be divisible by some prime $p$ with $p \leq k$ (for otherwise $n$ would be a nonempty product of primes each of which is at least $m$, so $n \geq m$). Then $p$ divides both $k!$ and $n$ so $k! + n$ is divisible by $p$ and has size at least $2p$, so is composite.
All this is to say that your argument is correct as far as it goes: what it establishes is the inequality
$N(k) \geq m -2$.
(In particular, since $m \geq k+1$, it follows that $N(k) \geq k-1$, a weaker but more commonly made observation.) Note that if $k = 2$, then the number of consecutive composites following $2! + 1 = 3$ is $1$: $4$ is composite, but $5$ is prime. Here $m = 3$ and $m-2 = 1$. Thus, your inequality is sharp when $k = 2$, i.e., it is an equality.
Now, as your computations show, it is usually the case that $N(k)$ is strictly larger than $m-2$, often much larger. This is not surprising: the above argument guarantees that the first $m-2$ numbers after $k! + 1$ will all be composite. After that it doesn't tell us anything. But most numbers are not prime: in a sense made precise by the Prime Number Theorem, if you choose a number randomly from the interval $[1,N]$, the chance it is prime is $\frac{1}{\log N}$, so if you choose a sufficiently large number at random, then the chance that it is prime is very close to zero, so we certainly expect several more values after $k! + (m-2)$ to be composite for most values of $k$. However, we are far from being able to predict exactly where the primes and composites will fall.
All this is to say that at least from an elementary perspective, I would be surprised if you were expected to say anything more than $N(k) \geq m -2$. (You didn't directly quote the statement of the problem, but whatever it is, I am guessing that this inequality, or perhaps even $N(k) \geq k-1$, is the intended answer.)
One could of course try to do better: for instance, surely there are infinitely many values of $k$ for which $N(k) > m-2$, and with sufficient knowledge and technique in analytic number theory one could probably prove that. (I'd be interested to see people try!) But I think it is exceedingly unlikely that there will be any kind of simple formula or expression for $N(k)$ in general, so you shouldn't be disappointed not to have found one!