What are the zero divisors of $\mathbb F_2[x]/\langle x^2+1\rangle$?
I think it is not a field, since the polynomial is reducible over $\mathbb F_2$. And because the ring is finite one has to show that it has a zero divisor, is it:
$$\mathbb F_2[x]/\langle x^2+1\rangle =\mathbb F_2[x]/\langle (x+1)^2\rangle $$
I'm stuck here, does the latter imply that $x^2=x=1$, so are there less than $4$ elements, otherwise I cannot find a zero divisor, can you help ?
It is true that, over the field with two elements, $x^2+1=(x+1)^2$.
However, it is false that $x^2=x$, because $x$ is intended to be an indeterminate or, in other words, a transcendental element over the field.
You can avoid ambiguities if you try with $F[x]/\langle(x-r)^2\rangle$, where $F$ is any field and $r\in F$.
Any element of the quotient ring can be written in a unique way as $a+bx+\langle(x+1)^2\rangle$, so there are four of them $$ 0+I,\quad x+I,\quad 1+I,\quad 1+x+I $$ where $I=\langle(x+1)^2\rangle$; it shouldn't be difficult to do the check.
For instance, $(1+x+I)^2=0+I$. Is $x+I$ a zero divisor?