Ok, so the basic idea whenever you have an extension which is the appendation of multiple elements is to take things one at a time. Namely, let's first find $n$ and $m$ in the following diagram
$$\begin{array}{c}F_p(X,Y)\\ \vert n\\ F_p(X^p,Y)\\ \vert m\\ F_p(X^p,Y^p)\end{array}$$
The basic idea now is that going from $F_p(X^p,Y^p)$ to $F_p(X^p,Y)$ is just appending one element since $F_p(X^p,Y)=F_p(X^p,Y^p)(Y)$. So, the question is what is the minimal polynomial of $Y$ over $F_p(X^p,Y^p)$? Well, we clearly have a candidate since $f(t)=t^p-Y^p\in F_p(X^p,Y^p)[t]$ and $f(Y)=0$. The question then is why $f(t)$ is irreducible in $F_p(X^p,Y^p)[t]$. The reason why would expect it to be is that any factorization should be one involving the fact that $t^p-Y^p=(t-Y)^p$--but of course this won't work since $t-Y\notin F_p(X^p,Y^p)[t]$. To make this more concrete suppose that $f(t)=p(t)q(t)\in F_p(X^p,Y^p)[t]$ is a factorization into irreducibles. Note then that we have factorized $f(t)$ in $F_p(X,Y)[t]$ and since $F_p(X,Y)$ is a field we know that $F_p(X,Y)[t]$ is a UFD and so use this to conclude that, up to constants, $p(t),q(t)$ are powers of $(t-Y)^k$. Conclude then that (from the primality of $p$ which tells you that $Y^k$ is not in $F_p(X^p,Y^p)$ for $k<p$) the one factor must be $(t-Y)^p$ and so the other is a unit. Conclude that $f$ is irreducible and so the minimal polynomial of $Y$ over $F_p(X^p,Y^p)$. Conclude that $m=p$.
Now you try for $n$ and then use the multiplicative property of field towers.
