Let $\left( \frac{\cdot}{p} \right)$ denote the Legendre symbol. We consider the two integers
$$M = \sum_{x=1}^{p-1} \left( \frac{1+x^4}{p} \right) \quad \text{and} \quad N = \sum_{x=1}^{p-1} \left( \frac{1+x^8}{p} \right).$$
Notice that $M$ and $N$ are integers with $|M|, |N| < p$. Also, we contend $M \equiv 4 \pmod 8$ and $N \equiv 0 \pmod 8$. For the sum $M$, there are exactly four residues $x$ with $x^4 = -1$, and the remaining residues make $2k-1$ groups of four contributing $\pm 4$ to the sum. Similarly, in the sum $N$ we have groups of eight contributing $0$ or $\pm 8$.
On the other hand modulo $p$ we have
$$M \equiv \sum_{x=1}^{p-1} (1+x^4)^{4k} \equiv (p-1) \left(2 + \binom{4k}{2k}\right) \pmod{p}$$
and
$$N \equiv \sum_{x=1}^{p-1} (1+x^8)^{4k} \equiv (p-1) \left(2 + 2\binom{4k}{k} + \binom{4k}{2k}\right) \pmod{p}.$$
whence
$$\frac{M-N}{2} \equiv \binom{4k}{k} \pmod{p}.$$
Note that $\frac{M-N}{2} \equiv 2 \pmod 4$ and lies in $(-p, p)$, hence
$$\frac{M-N}{2} \in \{\pm2, \pm6, \dots, \pm(p-3)\}.$$
Consequently $r \in \{2, 3, 6, 7, 10, 11, \dots, p-3, p-2\}$. In particular, since squares are $0$ or $1$ mod $4$, it follows $r$ is not a square.