= Solution
Use $R=\mathbb F_2[t]$. At $t=1$ the specialization is $X^5+X^2+1$, already proved irreducible and separable. Because the original <polynomial> is monic, this proves irreducibility over $\mathbb F_2(t)$ and gives a transitive group on five roots.
At $t=0$ the specialization factors as
$$
X^5+X+1=(X^2+X+1)(X^3+X^2+1).
$$
Neither factor has a root in $\mathbb F_2$, so they are irreducible; the factors are coprime and the specialization is separable. For example its derivative is $(X+1)^4$ and its value at one is one. The unramified reduction test gives a $(2)(3)$ <permutation> whose cube is a <transposition>. The same primitive prime-degree graph argument now gives
$$
\boxed{\operatorname{Gal}(f,\mathbb F_2(t))=S_5.}
$$
No characteristic-zero discriminant-sign argument is being transferred to characteristic two: good specialization and the explicit cycle test are sufficient.
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