$f(x) = x^p-s = x^p - t^p = (x-t)^P$
I don't know how to show that it is irreducible... How can I use Eisenstein to show this? Is it suitable to say by Gauss' lemma, since in the question $p$ is prime, $p=(t)$ and $t \notin p^2$ so it is irreducible?
If $t$ is transcendental over $\mathbb{F}_p$, then so is $t^p$. Thus, $\mathbb{F}_p(t^p)\cong\mathbb{F}_p(y)$ for a transcendental variable $y$ (I switch so we are not distracted by the exponent in "$t^p$"; we want to treat it as a transcendental variable, not the $p$th power of something.)
So... is $x^p-y$ irreducible over $\mathbb{F}_p(y)$? Yes. To see this, we can apply Gauss's Lemma. The polynomial $x^p-y$ is irreducible in $\mathbb{F}_p(y)[x]$ if and only if it is irreducible in $\mathbb{F}_p[y][x]=\mathbb{F}_p[x,y]=\mathbb{F}_p[x][y]$. But as a polynomial in $(\mathbb{F}_p[x])[y]$ it is clearly irreducible, since it is of degree $1$. Therefore, it is also irreducible over $\mathbb{F}_p(y)$.