We will show for any prime $p\ge 7$ there are two consecutive positive quadratic residues of $p$ that differ by $2$. The result is easy to verify for the prime $7$. So we suppose that $p\gt 7$.
For brevity we write QR for quadratic residue of $p$, and NR for quadratic non-residue of $p$.
Let $q$ be the smallest positive NR of $p$. Note that $q$ is prime. If $q$ is odd, then $q-1$ and $q+1$ are consecutive QR that differ by $2$, since all their prime factors are $\lt q$.
So we are finished unless $2$ is a NR. Assume $2$ is a NR. If $3$ is a QR, we are finished. So we may assume that $3$ is a NR.
Then $6$ is a QR, and therefore we are finished unless $5$ is a QR. If $7$ is a QR, we are finished, for $8$ is NR and $9$ is QR.
So we have solved our problem unless $2$ is NR, $3$ is NR, $5$ is QR, and $7$ is NR.
But in that case $14$ is QR, $15$ is NR, and $16$ is QR. The result follows.