The Chowdhury paper is available here. He proves:
Theorem. If $n>1$ and $2\le k\le n$, then either
$\qquad(1)\quad n!+k$ has a prime factor greater than $n$, or
$\qquad(2)\quad k$ is prime, $k>n/2$, and $n!+k$ is a power of $k$.
His argument is essentially as follows:
Proof. Suppose that $n!+k$ has no prime factor greater than $n$, and let $p\le n$ be a prime dividing $n!+k\,$; clearly $p\mid k$. Suppose first that $pn$. But then $q\mid k\left(\frac{n!}k+1\right)=n!+k$, a contradiction. Thus, $k$ must itself be prime, and $n!+k=k^m$ for some $m\ge 2$. Then $k\nmid k^{m-1}-1=\frac{n!}k$, so $k>\frac{n}2$. $\dashv$
He then cites T. Grundhöfer, Über die Zahlen der Form $n!+k$, Arch. Math. 33, 361-363 (1979), for the result that if $n>5$, none of the numbers $n!+k$ with $2\le k\le n$ can be a prime power. Thus, for $n\ge 6$ the second alternative of the Theorem cannot hold, and $n!+k$ must have a prime factor greater than $n$.
Added: I have now seen Grundhöfer’s paper. He first proves a
Lemma. For all $n\in\mathbb{Z}^+$, $n!\le 2\left(\frac{n}2\right)^n$.
The proof is by induction: $$2\le\left(1+\frac1n\right)^n=\left(\frac{(n+1)/2}{n/2}\right)^n\;,\tag{1}$$ whence $$\begin{align*} (n+1)!&=(n+1)n!\\ &\le(n+1)\left(2\left(\frac{n}2\right)^n\right)\\ &\le(n+1)\left(\frac{n+1}2\right)^n\tag{by (1)}\\ &=2\left(\frac{n+1}2\right)^{n+1}\;. \end{align*}$$
Then he proves that the only pairs $\langle n,k\rangle$ such that $n\ge 2$, $2\le k\le n$, and $n!+k$ is a prime power are $\langle 2,2\rangle,\langle 3,2\rangle,\langle 3,3\rangle,\langle 4,3\rangle$, and $\langle 5,5\rangle$, which are respectively $2^2,2^3,3^2,3^3$, and $5^3$.
Proof. Let $p$ be a prime such that $n!+k=p^r$, where $n\ge 2$ and $2\le k\le n$. Since $k\mid n!+k$, $k=p^s$ for some $s$ with $1\le s$$n!=p^r-p=p(p^{r-1}-1)\;.\tag{2}$$ Moreover, we must have $p>n/2$. When $p=2$, the righthand side of $(2)$ is congruent to $2$ mod $4$, so $n<4$, and we have the first two exceptional $\langle n,k\rangle$ pairs, $\langle 2,2\rangle$ and $\langle 3,2\rangle$. Assume henceforth that $p$ is an odd prime.
For $a\in\mathbb{N}$ let $m(a)$ be the largest $m$ such that $$2^m\mid p^{2^a}-1=\left(p^{2^{a-1}}-1\right)\left(p^{2^{a-1}}+1\right)\;.$$ The factorization shows that $m(a+1)=m(a)+1$ for $a\ge 1$, so $m(a)=m(1)+a-1$ for $a\ge 1$. Moreover, $m(0)\le m(1)-1$.
Write $r-1=2^au$, where $u$ is odd. Then $$p^{r-1}-1=\left(p^{2^a}-1\right)\sum_{i=0}^{u-1}p^{2^ai}\;,$$ so $m(a)$ is the highest power of $2$ dividing $p^{r-1}-1$. Equating the maximum powers of $2$ dividing the two sides of $(2)$ yields $$\sum_{i\ge 1}\left\lfloor\frac{n}{2^i}\right\rfloor=m(a)\le m(1)+a-1\tag{3}\;.$$ Note that since $2^{m(1)}$ is the highest power of $2$ dividing $p^2-1=(p+1)(p-1)$, we must have $m(1)\le\lg(p+1)+1\le\lg(n+1)+1$, where $\lg$ is $\log_2$.
Next, note that by the lemma we have $p^r-p=n!\le 2\left(\frac{n}2\right)^n<2p^n$ (recall that $p>n/2$), which is false for $r>n$, so $r\le n$. Since $r-1=2^au\ge 2^a$, this implies that $$a\le\lg(r-1)\le\lg(n-1)\;,$$ whence from $(3)$ we have $$\sum_{i\ge 1}\left\lfloor\frac{n}{2^i}\right\rfloor=m(a)\le \lg(n+1)+1+\lg(n-1)-1<2\lg n\;.\tag{4}$$ I claim that $(4)$ has only finitely many solutions.
Clearly $$\sum_{i\ge 1}\left\lfloor\frac{n}{2^i}\right\rfloor\ge\left(\frac{n}2-\frac12\right)+\left(\frac{n}4-\frac34\right)=\frac14(3n-5)\;,$$ so $(4)$ implies that $3n-5<8\lg n$, i.e., $3n-8\lg n<5$. The function $3n-8\lg n$ is monotone increasing for $n\ge 4$, and for $n=11$ we already have $$33-8\lg 11=33-4\lg 121>33-4\lg 128=5\;,$$ so $(4)$ implies that $n\le 10$ and thence that $$\sum_{i\ge 1}\left\lfloor\frac{n}{2^i}\right\rfloor\le\lfloor 2\lg 10\rfloor=\lfloor\lg 100\rfloor=6\;.$$ Thus, we must have $n\le 7$. Since $p>n/2$ and $p\ge 3$, the only possible $\langle n,k\rangle$ pairs are $\langle 3,3\rangle,\langle 4,3\rangle,\langle 5,3\rangle,\langle 5,5\rangle,\langle 6,5\rangle,\langle 7,5\rangle$, and $\langle 7,7\rangle$, of which only the first, second, and fourth are prime powers. $\dashv$