Are there any $n$ for which $ n^4+n^3+n^2+n+1$ is a perfect square?

Are there any positive $n$ for which $ n^4+n^3+n^2+n+1$ is a perfect square?

I tried to simplify

\begin{align*} n^4+n^3+n^2+n+1 &= n^2(n^2+1)+n(n^2+1)+1\\ &= (n^2+n)(n^2+1)+1 \\ &= n(n+1)(n^2+1)+1 \end{align*} Then I assumed that the above expression is a square; then

$$ n(n+1)(n^2+1)+1 = k^2$$

$$ \begin{align*} n(n+1)(n^2+1) &= (k^2-1) \\ &= (k+1)(k-1) \end{align*} $$

Then trying to reason with prime factors, but cannot find a concrete proof yet.

Assuming you want positive integers $n$,

I believe we can show that

$$(2n^2 + n)^2 \lt 4(n^4 + n^3 + n^2 + n + 1) \lt (2n^2 + n + 1)^2$$

for $n \gt 3$.

Note: A similar inequality can be given for negative $n$.

Here is my answer!

H. Bensom, Germany