At a more abstract elevation, we can view this particular case as a corollary to more general facts from commutative algebra that I expect can be found in most introductory notes or texts.
Lemma 1. If $R$ is an integral domain, then the polynomial ring $R[x]$ is also an integral domain.
Proof. Let $ab=0$ in $R[x]$. Write $a=a'x^n+\cdots$ and $b=b'x^m+\cdots$. Then we have
$ab=a'b'x^{n+m}+\cdots=0\implies a'b'=0\implies a'\text{ or }b'=0.$
This however contradicts the implicit supposition that $\deg a=n$ and $\deg b=m$.
Lemma 2. If $K$ is a field, then the polynomial ring $K[x]$ is a principal ideal domain.
Proof. Let $G=(g_1,\dots,g_n)$ be an ideal of $K[x]$. By induction on Bezout's identity, there must exist polynomials $a_1,\dots, a_n$ such that $a_1g_1+\cdots a_ng_n=d=\gcd(g_1,\dots,g_n)$. Since each $g_i$ is a multiple of $d$, every $K[x]$-linear combination of the $g_i$'s is a multiple of $d$ and conversely, since $d\in G$ and $G$ is an ideal, every multiple of $d$ is in $G$. Therefore $G=(d)$, which is principal. By Lemma 1, we also know that $K[x]$ is a domain, so it is a principal ideal domain.
Lemma 3. If $K$ is a field and $f\in K[x]$ is irreducible, the ideal $(f)$ is maximal.
Proof. Suppose $(f)\subsetneq G\subsetneq K[x]$. By Lemma 2, $G=(g)$ for some $g$. Now $f\in(g)$ implies $f=ag$ for some $a$; if $a$ is a unit then $g=a^{-1}f\in(f)$ implies $(g)\subseteq(f)$, contradicting our hypothesis, or else $f$ factors nontrivially, which contradicts our supposition that $f$ is irreducible.
Lemma 4. If $R$ is an integral domain and $I$ a maximal ideal, then the factor ring $R/I$ is a field.
Proof. Let $a + I\in R/I$ be nonzero. Since $\{i+ra:i\in I,r\in R\}$ is an ideal in $R$ containing $I$ and $a$, it must be all of $R$, whence $i+ra=1$ for some $i\in I,r\in R$, ergo $(r + I)(a+I)=1+I$. Since the nonzero cosets have inverses, the factor ring must be a field.
Now $\mathbb{F}_2=\mathbb{Z}/2\mathbb{Z}$ is a field, hence $\mathbb{F}_2[x]$ is a PID. The polynomial $f=x^2+x+1$ is irreducible, so $(f)$ is maximal, from which we conclude $\mathbb{F}_2[x]/(f)$ is a field. For more on commutative algebra, I recommend the Algebra Handouts in the Number Theory I course or the Commutative Algebra notes from Pete L. Clark (the latter is much more advanced).