Show that $X^4+TX^2+T$ is irreducible in $ F_{2}(T) $
Using Eisenstein with T as a prime element this proof is simple. Can I proof that T is prime any easier than in the folowing:
Theorem 1: K is PID then
p is irreducible $\Leftrightarrow$ p is prime element $\Leftrightarrow$ (p) is maximal ideal
Since $ K=F_{2}(T) $ is a field $ K[X] $ is a principal ideal domain (PID).
Assume (T) is not an maximal ideal. Obviously (T) contains all polynomials P with $deg_{T}(P) \geq 1$. So in order to find (m) $\supsetneq$ (T) with $(m) \neq K[X]$ we have to add an element with $deg_{T}(P) = 0$ to (T). Let $a \neq 0 \in F_{2}$ be this element. Because a is a unit we have $ 1 \in (T)$ Thats a contradiction to $(m) \neq K[X]$.
$\Rightarrow$ (T) is an maximal ideal and therefore (Theorem 1) T is prime.
I guess the full argument (which you probably imply) should be the following.
By Gauss' Lemma, it is enough to show that the polynomial $$ f = X^4+TX^2+T $$ is irreducible in $R[X]$, where $R = F_{2}[T]$.
Now you indeed use Eisenstein's Lemma, and show that $T$ is prime in $R$. For this, is it enough to note that a polynomial of degree $1$ over a field is always irreducible.