Let $m_t(p)$ be the least integer such that for any subset $S \subset \mathbb F_p$ with $|S|\geq m_t(p)$ there exist $x,y\in S$ such that $xy+t-1$ is a square in $\mathbb F_p$.
Then since the map $x\mapsto x^2-1$ is two-to-one, we have $n_t(p) \leq 2 m_t(p)$.
If $S$ has the property that there does not exist $x,y\in S$ such that $xy+t-1$ is a square then we have $$- |S|^2= \sum_{x,y\in S} \left( \frac{xy+t-1}{p} \right) = \sum_{\substack{ x,y\in S\\ x=0\textrm{ or }y=0}} \left( \frac{t-1}{p} \right) + \sum_{x,y\in S\setminus \{0\} } \left( \frac{xy+t-1}{p} \right)$$
$$ = \sum_{\substack{ x,y\in S\\ x=0\textrm{ or }y=0}} \left( \frac{t-1}{p} \right) + \frac{1}{p-1} \sum_{x,y\in S\setminus \{0\} } \sum_{\chi \colon \mathbb F_p^\times \to \mathbb C^\times} \chi(xy)
\sum_{ u\in \mathbb F_p^\times} \overline{ \chi(u)} \left( \frac{u+t-1}{p} \right)$$
Now $\sum_{ u\in \mathbb F_p^\times} \overline{ \chi(u)} \left( \frac{u+t-1}{p} \right)$ is a Jacobi sum and bounded by $\sqrt{p}$ (for $t\neq 1$) and so
$$ \left| \frac{1}{p-1} \sum_{x,y\in S\setminus \{0\} } \sum_{\chi \colon \mathbb F_p^\times \to \mathbb C^\times} \chi(xy)
\sum_{ u\in \mathbb F_p^\times} \overline{ \chi(u)} \left( \frac{u+t-1}{p} \right) \right| \leq \frac{\sqrt{p}}{p-1} \sum_{\chi \colon \mathbb F_p^\times \to \mathbb C^\times} \left| \sum_{x,y\in S\setminus \{0\} } \chi(xy)\right|$$
$$= \frac{\sqrt{p}}{p-1} \sum_{\chi \colon \mathbb F_p^\times \to \mathbb C^\times} \left| \sum_{x S\setminus \{0\} } \chi(x)\right|^2= \sqrt{p} ( |S\setminus\{0\}| \leq \sqrt{p} |S|$$
while the first terms is bounded by $2 |S|-1 < 2 |S|$ so we get
$$ |S|^2 <\sqrt{p} |S|+ 2|S|$$
and thus $$ |S|< \sqrt{p}+2$$
i.e. we have $m_t(p)\leq\sqrt{p}+2$ and $n_t(p) \leq 2\sqrt{p}+2$.
