Trying to paint a coherent picture while hopefully also answering the question.
Here we use two different polynomials in defining the field $GF(2^{233})$, namely
$$f_1(z)=z^{233}+z^{74}+1\qquad\text{and}\qquad f_2(z)=z^{233}+z^{159}+1.$$
They are both irreducible. Actually it suffices to verify that one is irreducible, because they are each others reciprocal polynomials. That is,
$$
z^{233}f_1(\dfrac1z)=f_2(z).\tag{1}
$$
With these two polynomials we can define two variants of $GF(2^{233})$. Namely the fields
$$K_1=GF(2)[z]/\langle f_1(z)\rangle\qquad\text{and}\qquad K_2=GF(2)[z]/\langle f_2(z)\rangle.$$
By the fundamental theorem of finite fields we know that they are isomorphic. The isomorphism is by no means unique (there are $233$ different automorphisms to choose from), but one of them stands out because of $(1)$. If we denote the natural generators $\alpha=z+\langle f_1(z)\rangle\in K_1$ and $\beta=z+\langle f_2(x)\rangle\in K_2$, then, all because of $(1)$, we have an isomorphism $\sigma:K_1\to K_2$ uniquely determined by $\sigma(\alpha)=1/\beta$. This is because $(1)$ says that $1/\beta$ is a root of $f_1(z)$ as is $\alpha$, and an isomorphism of fields must observe such polynomial relations.
If we look at an elliptic curve
$$E:y^2+a_1 xy+a_3 y=x^3+a_2 x^2+a_4 x+a_6,\tag{2}$$ where $a_1,a_2,a_3,a_4,a_5,a_6\in K_1$, then we can think of the "same" curve as being defined over $K_2$, if we apply the isomorphism $\sigma$ everywhere. We end with
$$
E':y^2+a_1' xy+a_3' y=x^3+a_2' x^2+a_4' x+a_6',\tag{2'}
$$
where $a_i'=\sigma(a_i)\in K_2$ for all indices $i$. In other words, we replace the coefficients $a_i\in K_1$ with their isomorphic images in $K_2$.
As isomorphisms of fields respect the arithmetic operations, it immediately follows that if a point $P=(x,y)\in K_1\times K_1$ lies on the curve $E$, then $P'=(x',y')\in K_2\times K_2, x'=\sigma(x), y'=\sigma(y)$, is a point on the curve $E'$.
Furthermore, field automorphisms also take lines in $K_1\times K_1$ to lines in $K_2\times K_2$, and this implies that the above mapping (still call it $\sigma$) also takes the addition of $E$ to addition of $E'$, so it is automatically also an isomorphism of the underlying groups of the two elliptic curves. So if $k$ is an integer and $Q=k*P=(u,v)\in E$ is an integer multiple of $P$, then $Q'=k*P'=(u',v')$ where $u'=\sigma(u),v'=\sigma(v)$.
An isomorphism between the underlying fields automatically produces an isomorphism of elliptic curves and their group structures provided that you also apply the isomorphism to the coefficients of the defining equation (like the passage from $E$ to $E'$ above).
Recording the following, just in case. Putting on my algebra teacher's hat :-). A mistake often made by people not well versed in the language of quotient rings of polynomial rings is to equate the coset $z+\langle f_1(z)\rangle$ with the polynomial $z$. Thinking that $z$ could be an element of $K_1$. The following confusion then rears its ugly head. This element is totally unrelated to the element $z+\langle f_2(z)\rangle\in K_2$. The reason I denoted them by $\alpha$ and $\beta$ respectively is exactly to avoid this confusion.
It is sometimes convenient to denote the coset of $z$ by $z$ as well, but you can only do this if the field description never changes. Compare with modular arithmetic. Modulo $11$ the coset of $2$ (similarly often just denoted $2$) really is
$$\overline{2}=\{2,13,24,35,\ldots,-9,-20,-31,\ldots\}$$
but "the same" coset of $2$ modulo $13$ looks like
$$\overline{2}=\{2,15,28,41,\ldots,-11,-24,-37,\ldots\},$$
a totally different animal. It's the same thing with cosets of polynomials.
Caveat: More often than not when there are two alternative definitions of a finite field, the relation between the respective zeros of the two polynomials is more complicated. The case of reciprocal polynomials here is very exceptional. I simply could not resist using it.
