Is $(7,4)$ the only non-trivial integer solution for $(n)_k=n!$?

Question

I accidentally noticed that:

$$(7)_4=7 \cdot 8 \cdot 9 \cdot 10=2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7=7!$$

Here $(n)_k$ is the Pochhammer symbol.

I wonder, are there any other non-trivial integer solutions $(n,k)$?

$$(n)_k=n!$$

Among the ones I consider trivial we have $(0,0),(1,0),(1,1),(2,1)$. Somehow, I'm sure that I will get a lot of comments with these four solutions.

This is the implicit plot of the equivalent equation:

$$\Gamma (n+k)=n \Gamma^2 (n)$$

Answer 1

If $n+k\geq 14$, then there are at two primes in the interval $\left(\frac{n+k-1}{2},n+k-1\right]$. Therefore, $(n+k-1)!$ cannot equal $n\cdot \big((n-1)!\big)^2$. This leaves the cases $n+k\leq 13$ to be dealt with.