$n^3+n+1$ is perfect square find all positive integers $n$ for which $n^3+n+1$ is a perfect square
$n^3+n+1$ is odd so $8|n^3+n$ so $8|n(n^2+1)$ if $2|n^2+1$  ,$v_2(n^2+1)=1$ so $n$ must be even, contradiction. So $8\mid n.$
Also by checking $\bmod7$ we understand that $n=0\pmod7$ or $n=2\pmod7$ so $56\mid n$ or $n=16\pmod{56}$ also $72$ is an answer!
Another approach: add $n^2$ to both sides: $(n+1)(n^2+1)=a^2+n^2$ It's obvios that $\gcd(a,n)=1$ so by vitek lemma all prime divisors of $a^2+n^2$ are $4k+1$ so the same for prime divisors of $n+1$.
Another approach: let $[\sqrt{n^3}]=c$
So $n^3<(c+1)^2$ so $n+1<c$ so $n^3+n+1<(c+2)^2$ so if $n^3+n+1=a^2$ $a$ is less than or equal to $c+1$ on the other hand it's obvious that $a>c$ so $a=c+1$

The solutions correspond to the integral points on the elliptic curve
$$
y^2=x^3+x+1,
$$
which are given by
$$
(0,\pm 1),\;(72,\pm 611).
$$
See for example LMFDB.
It is easy to obtain the points $(72,\pm 611)$ starting from $(0,1)$, see
Find an affine point on $E:y^2=x^3+x+1$
but it requires more on elliptic curves to show that there are no other integer points.
