Note that $f(x)=x^2+1=(x+1)^2$ has distinct two roots $1,\ i$ so that it is separable polynomial.
Hence spliting field is $K={\bf Z}_2(i)=\{ 0,\ 1,\ i\}$
Over $K$, $f(x)$ has factorization $(x-1)(x-i)=x^2+x+ix+1\neq f(x)$
Why does such phenomenon happen ?
More detailed explanation :
Consider ${\bf R}$-case. For $f(x)=x^2+1 \in {\bf R}[x]$ has a root $\pm i$ so that we have an extension $$ K={\bf R}(i)={\bf C}={\bf R}[x]/(x^2+1)$$ and we have factorization $$ f(x)=(x+i)(x-i)$$ over $K$.
I want to do similar thing in case $F={\bf Z}_2$