What do the elements of the field $\mathbb{Z}_2[x]/(x^4+x+1)$ look like? What is its order?

The elements of $F$ are $\{ f(x) + (x^4 + x + 1) \mid f(x) \in \mathbb{Z}_2[x], \deg f < 4 \}$. There are $2^4$ of them. Any field of order $2^4$ is isomorphic to $F$.

In general, if $p(x) \in \mathbb{Z}_2[x]$ is irreducible of degree $k$, then $\mathbb{Z}_2[x]/(p(x))$ is a field of order $2^k$.

There is a notation that makes this field more convenient to work with. Let $\alpha = x + (x^4 + x + 1) \in F$. Then for $f(x) \in \mathbb{Z}_2[x]$, $f(\alpha) = f(x) + (x^4 + x + 1)$. So, for example, we can write the element $x^2 + 1 + (x^4 + x + 1)$ as $\alpha^2 + 1$. In this notation,

$$F = \{ f(\alpha) \mid f(x) \in \mathbb{Z}_2[x], \deg f < 4 \}.$$

An isomorphic field is the nimber field of nimbers less than 16. The representation of the elements is simpler, but I'm finding nim-multiplication to be harder than polynomial multiplication (maybe there's a trick to it that I don't know).

In this ring, $x^4+x+1 = 0$, or equivalently $x^4=x+1$. So $F=\{ax^3+bx^2+cx+d \mid a,b,c,d\in\mathbb{Z}_2\}$. Any higher powers can be reduced using $x^4=x+1$. Note that this means that this field is isomorphic to the finite field of 16 elements.

You were asking for other fields that are isomorphic to $F=\Bbb{Z}_2[x]/\langle x^4+x+1\rangle=\Bbb{Z}_2[\alpha]$ with $\alpha=x+\langle x^4+x+1\rangle$. I proffer one such. Unfortunately it doesn't easily generalize to all finite fields. Thus it is a bit ad hoc.

Consider the ring $R=\Bbb{Z}[\zeta]$, where $\zeta=e^{2\pi i/5}$ is a primitive fifth root of unity. In $\Bbb{Z}_2[x]$ we have $$ x^{12}+x^9+x^6+x^3+1=(x^4+x+1)(x^8+x^4+x^2+x+1). $$ Therefore the element $\beta=\alpha^3\in F$ satisfies the equation $\beta^4+\beta^3+\beta^2+\beta+1$. As this coincides with the minimal polynomial of $\zeta$ there exists a homomorphism of rings $f:R\to F$ determined by $f(\zeta)=\beta$.

This homomorphism is surjective. One way of seeing that follows from the calculation $$ \begin{aligned} \alpha&=\alpha\cdot(1+0)^2=\alpha\cdot(1+1+\alpha+\alpha^4)^2\\ &=\alpha\cdot(\alpha+\alpha^4)^2=\alpha\cdot(\alpha^2+\alpha^8)\\ &=\alpha^3+\alpha^9=\beta+\beta^3. \end{aligned} $$ What about $\ker f$? Clearly $2R\subseteq \ker f$. Because $R$ is a free abelian group of rank four, we know that $|R/2R|=2^4=16$. This forces $2R=\ker f$.

Conclusion: $$ F\cong R/2R=\Bbb{Z}[\zeta]/\langle 2\rangle. $$