We consider three cases depending on the value of $p$.

Case 1. $p \equiv 1 \pmod 4$.

Picking $t = 2$ works. Since $x^2$ and $y^{\frac{p-1}{2}}$ are perfect squares, their difference cannot be $2$.

Case 2. $p \equiv 3 \pmod 4$ and $p > 7$.

Note that $y^{\frac{p-1}{2}} \equiv -1, 0$ or $1 \pmod p$. Thus, it suffices to show that there exists $t$ such that $t-1, t, t+1$ are quadratic nonresidues modulo $p$. To this end, note that the squares $1, 25, 49$ form an arithmetic progression with common difference $24$. Let $c$ be the inverse of $24$ modulo $p$, then
$$c+1 \equiv c(1+24) \equiv 25c \pmod p, c+2 \equiv c(1+48) \equiv 49c \pmod p,$$
which implies $\left(\frac{c}{p}\right) = \left(\frac{c+1}{p}\right) = \left(\frac{c+2}{p}\right)$. So when $\left(\frac{c}{p}\right) = -1$, picking $t \equiv c+1 \pmod p$ works.
On the other hand, when $\left(\frac{c}{p}\right) = 1$, then since $p \equiv 3 \pmod 4$ we have $\left(\frac{-1}{p}\right) = -1$, so picking $t \equiv -c-1 \pmod p$ works.

Case 3. $p=7$. We will show that $t=7$ works, i.e. that the equation
$$x^2 = y^3 + 7$$
has no integer solution. Suppose the contrary. If $y$ is even, then $x^2 \equiv 3 \pmod 4$, which is impossible. Thus $y$ is odd. Write the equation as $x^2 + 1 = (y+2)(y^2 - 2y + 4)$, and note that $y^2 - 2y + 4 = (y-1)^2 + 3 \equiv 3 \pmod 4$. So there exists a prime $q \equiv 3 \pmod 4$ such that $q \mid y^2 - 2y + 4$.
However, since $q \mid y^2 - 2y + 4$, so $q \mid x^2 + 1$, and thus $\left(\frac{-1}{q}\right) = 1$, which contradicts with $q \equiv 3 \pmod 4$. Therefore the equation has no integer solutions.