I claim $n=1$ is the only answer. Verify $n\le 3$ manuallly; let $n>3$. Assume first $2n+3$ is not a prime, then there is a $d\mid 2n+3$ such that $d\le (2n+3)/3$. As $(2n+3)/3\le n$ for $n\ge 3$, it follows $d\mid n!$. This is absurd. Now suppose $2n+3$ is a prime. We have (a) $n!\equiv 1\pmod{2n+3}$ and (b) $(2n+2)!\equiv -1\pmod{2n+3}$ due to Wilson. Studying the second equation, \[ (2n+2)! = n! (n+1)(n+2)(n+3)\cdots (2n+2)\equiv 1\cdot (n+1)(n+2)(-1)^n n!\equiv (-1)^n(n+1)(n+2)\pmod{2n+3}. \]Setting $n\equiv -3/2\pmod{2n+3}$ then inserting, the rest is easy.

Checking $n=1,2,3$ yields $n=1$ is answer...Assume $n>3$ Claim1: $2n+3 \text{ is a prime number }$ Let $p$ be arbitrary prime divisor of $2n+3$.If $p<n \implies p \nmid n!-1 \implies \text{contradiction}$. Hence prime divisors of $2n+3$ is greater than $n$.Assume $p_1$ and $p_2$ are different prime divisors of $2n+3$ Then $2n+3=p_1p_2>n^2 \implies \text{contradiction}$.Thus $2n+3$ is a prime $\square$ Claim2: $(n+1)(n+2) \equiv 1 \text{ or} -1 \pmod {(2n+3)}$ Note that $n! \equiv 1 \pmod {(2n+3)}$. Since $2n+3$ is prime , Wilson theorem yields $$(2n+2)! \equiv n!(n+1)...(2n+2) \equiv (n+1)...(2n+2) \equiv (n+1)(n+2)(-1)^nn! \equiv (n+1)(n+2)(-1)^n \equiv -1\pmod {(2n+3) } \square$$ Claim3: $\text{There is not answer if } n>3$ $n$ is even $$(n+1)(n+2) \equiv n^2+3n+2 \equiv n^2+n-1 \equiv -1 \pmod {(2n+3)} \iff 2n+3|n \text{ or } 2n+3|n+1 \implies \text{ no answer}$$$n$ is odd $$(n+1)(n+2) \equiv n^2+3n+2 \equiv n^2+n-1 \equiv 1 \pmod {(2n+3)} \iff 2n+3|n+2 \text{ or } 2n+3|n-1 \implies \text{ no answer } \square$$ so we are done

$n=1$ is the only answer, by bashing Euclidean algorithm and showing that $(2n+3, n!-1) = 1 \quad \forall n > 1$ .

It is easy to see that all prime divisors of $2n+3$ are greather strictly than $n$ If $2n+3$ has two prime divisors $p,q$ then $pq\mid2n+3 \implies n^2<pq\le2n+3$ contradiction And similarly if $p^2\mid2n+3 \implies n^2<p^2\le2n+3$ contradiction Hence $2n+3$ is a prime number let $2n+3=p$ Then $$ \left(\frac{p-3}{2}\right)! \equiv 1\pmod p \iff \left(\frac{p-1}{2}-1\right)!\equiv 1 \pmod p \iff \left(\frac{p-1}{2}\right)!\equiv \frac{p-1}{2} \pmod p$$ It is easy to see that $\forall k \in \{1,\ldots,p-1\}$ $(p-k)!\equiv (-1)^k \frac{1}{(k-1)!} \pmod p$ just from wilson's theorem Setting $k=\frac{p+1}{2}$ we get $$\left(\left(\frac{p-1}{2}\right)!\right)^2\equiv (-1)^\frac{p+1}{2} \pmod p$$ Hence we get $$\left(\frac{p-1}{2}\right)^2\equiv \pm 1 \pmod p \implies \pm4 \equiv 1 \pmod p \implies p=5 \quad or \quad p=3$$ Hence $2n+3=5$ since $2n+3>3$ Thus $n=1$ is the only solution $Q.E.D$

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I claim $n=1$ is the only answer. If $2n+3$ is not a prime, then consider $p | 2n+3$. But because $2n+3$ is odd, $p \leq \frac{2n+3}{3} \leq n$, so $p|n!$. Therefore, $2n+3|n!-1$ is absurd by considering modulo $p$.

If $2n+3$ is a prime, we need to prove that $n! \equiv 1 \pmod{2n+3}$ and $(2n+2)! \equiv -1 \pmod{2n+3}$ is impossible by Wilsons. Writing the latter $n$ terms of $(2n+2)!$ "backwards", the latter equivalence turns into $$n! (n+1)(n+2) \cdot (-1)(-2) \dots (-n) \equiv -1 \pmod{2n+3}$$$$(-1)^n (n+1)(n+2) \equiv -1 \pmod{2n+3}$$Do casework on if n is even[hide]$n^2+3n+3 \equiv 0 \pmod{2n+3}$. Euclidean algorithm gives $3 \equiv 0 \pmod{2n+3}$, no solutions[/hide] or n is odd

finishes.