Assume by contradiction. Let $n^3 - n - 3 = k^2$. The equation is equivalent to $(n+2)(n^2 - 2n + 3) = k^2 + 9$. (1) $n$ is even. From the original equation we have $n \equiv 4$ (mod 8) Then $n+2$ is of the form $8k+6$, thus yielding a prime divisor of the form $4k+3$. Observing RHS, the prime divisor must be 3, and $LHS$ has two $3$'s. But then $n+2$ and $n^2 - 2n + 3$ cannot be simultaneously multiples of $3$, and $n+2$ is hence a multiple of $9$. Then $\frac{n+2}{9}$ must yield a prime divisor of the form $4k+3$ again, contradiction. (2) $n$ is odd. $n+2$ must be of the form $4k+1$. This in turn makes $n^2 - 2n + 3$ of the form $8k+6$. Similar contradiction. Pretty well-known technique. Nice problem.

I would like to rewrite parts of the above proof and to fill in the missing details. I shall also offer my solution to $(2)$. Assume for a contradiction, i.e. there exist integers $n$ and $k$ such that $n^3-n-3=k^2$. Equivalent to $(n+2)(n^2-2n+3)=k^2+9$. Note that if $p$ is prime and $p$ is a factor of $k^2+9$, then $p=2, 3$ or $4|p-1$. (*) $(1)$ $n$ is even If $2^1$ is the greatest power of $2$ that divides $n$, $4|k^2-3$, yielding a contradiction. If $n$ is divisible by $8$, $k^2$ must be odd, so $8|k^2-1$, but $k^2=n^3-n-3 \equiv{5} \pmod{8}$, yielding a contradiction. Thus $2^2$ is the greatest power of $2$ that divides $n$. Hence $n+2$ is of the form $8l+6$ so $n+2$ is divisible by $3$ (by (*)). $4|n$ implies that $n^2-2n+3 \equiv{3} \pmod{4}$, so by (*) again, $n^2-2n+3$ is divisible by $3$. It is impossible for both $n+2$ and $n^2-2n+3$ to be divisible by $3$, yielding a contradiction $(2)$ $n$ is odd We obtain $(n-1)n(n+1) \equiv{4} \pmod{8}$ Both $n-1$ and $n+1$ are even and one of them is divisible by $4$, yielding a contradiction. As we have obtain a contradiction in all cases, the desired result holds.

Here my solution: Assume that $ n^3-n-3=k^2 $,then $ (n+2)((n-1)^2+2)=k^2+3^2 $. Case 1: Let $ n $ is odd.We have $ n^3-n-3=k^2=1 (mod 8) $, so $ n^3-n=4 (mod 8) $, but for any odd integer $ n $, we have $ n^3-n=0 (mod 8) $, so there are no solution in this case. Case 2: Let $ n $ is odd.Then $ n=4 (mod 8) $, so $ n+2 $ has prime divisor from the set $ P_3 $.Hence this prime is $ 3 $ and $ 3 | k (*) $.( we know that $ v_p(a^2+b^2) $ is even for $ p \in P_3 $ ).So $ n=1 (mod 3) $, and then $ (n-1)^2+2=2 (mod 3) $, so that $ 9 |n+2 \implies n=7 (mod 9) $.Then we have, $ k^2=7 (mod 9) $.But from $ (*) $ we know that $ 9 | k^2 $, a contradiction.

Assume for contradiction that $n^3-n-3=k^2$. Since $n^3-n-3=(n-1)n(n+1)-3$, we can conclude that $3|k$. Let $k = 3m$ and our equation becomes $n^3-n=9m^2+3$. Considering this equation modulo $8$, we have that $n\equiv 4\pmod{8}$ since quadratic residues modulo $8$ are $0,1,4$. Adding $6$ to both sides, $$(n^2-2n+3)(n+2)=9(m^2+1).$$Since $n\equiv 4\pmod{8}$, the LHS has factors that are both $3\pmod 8$ and $6\pmod 8$. Hence each factor on the LHS has at least one prime factor that is $3\pmod 4$ and $3$ cannot divide both factors so we have at least one prime factor $p\equiv 3\pmod{4}$ and $p>3$ that divides both sides of the equation. However this implies that $p|m^2+1$, a contradiction.

Suppose we can find $x$ such that $n(n-1)(n+1)=x^2+3$ If $p|n(n-1)(n+1)$ we have $x^2\equiv{-3}\pmod{p}$ With quadratic residues we understand that either $p=3$ or $p\equiv1\pmod{3}$ But it's obvious that one of the $n , n-1 , n+1$ is equal $2\pmod{3}$ so it has a prime divisor such that it's equal $2\pmod{3}$ , a contradiction $\blacksquare$

@above I did the same thing but that argument is slightly flawed, because $p$ has to be odd for this to work. Taking $\mod 8$ we see that $n\equiv 4\mod 8$. Then if $n-1$ or $n+1$ is $2\mod 3$, then we get a prime $q\equiv 2\mod 3$, if $n\equiv 2\mod 3$, then $\frac{n}{4}$ is odd and $2\mod 3$, so we get a such $q$ anyway.