Is there a constant $M \in \mathbb{N}$ so that for all primes $p \geq M$, $k \geq 2, k \in \mathbb{N}$ there exists a prime $q_{p,k} \geq p$ for which $q_{p,k}$ divides $\frac{(k+p-2)!}{(k-1)!}$?
If we temporarily allow $k = 1$ in the question then the answer is no as $(p-1)!$ is not divisible by any prime that is at least $p$. This question is clearly true for the prime $2$. This is true for the prime $3$ as $(k+1)(k)$ is never a power of $2$ when $k \geq 2$ (as one of $k,k+1$ is odd and $\geq 3$).
Note that we have the following; For all primes $p \geq 3$, $k \geq \text{lcm}(1,...,p-1)$ there exists a prime $q_{p,k} \geq p$ for which $q_{p,k}$ divides $\frac{(k+p-2)!}{(k-1)!}$. This can be figured out with set theory.
If $2 \leq k \leq p$ then $p \not | (k-1)!$ and $p| (k+p-2)!$ since $ k+p-2 \geq 2+p-2 = p$ and so $ p | \frac{(k+p-2)!}{(k-1)!}$ and we can take $q_{p,k} = p$
If $ k>p>p-1$ then $\frac{(k+p-2)!}{(k-1)!} = k(k+1) \cdots (k+p-2)$ and using sylvester-schur theorem we conclude that there is a prime divisor $q$ such that $q >p-1 $ and so $q\geq p$ and we are done.