Solving $n^5+n^4-3=x^2\pmod p$

Prove that for every odd prime number $p$ there is a natural number $n$ such that the equation $n^5+n^4-3=x^2\pmod p$ has no solutions.

So we have to understand that for each $p$ we can find $n$ such that the Legendre symbol $\left(\dfrac{n^5+n^4-3}{p}\right) = -1$. For $p=4k+3$ we can take $n=1$. How to deal with $4k+1$ case?

The curve $x^2=n^5+n^4-3$ has genus $2$, and so the Hasse-Weil bound implies that the number of ordered pairs $(x,n)$ modulo $p$ satisfying the equation is between $p-4\sqrt p$ and $p+4\sqrt p$. There can be at most five values of $n$ for which the right-hand side is congruent to $0\pmod p$; for all other values of $n$, if $x$ satisfies the congruence then so does $-x$. Therefore the number of distinct $n\pmod p$ that can appear is at most $5 + \frac12(p+4\sqrt p-5)$. This quantity is less than $p$ as soon as $p\ge29$, and so there exists $n\pmod p$ such that the equation has no solutions. And the result can be checked by hand for $3\le p\le 23$.