First of all, I want to clarify the terms. The natural meaning of "different" here is non-isomorphic - it has nothing to do with sets of axioms. And I'll take Henle and Kleinberg's definition of "hyperreal number system": a structure elementarily equivalent to $\langle \mathbb{R}, +,\cdot,0,1,<\rangle$ and containing both $\mathbb{R}$ and an infinitesimal element (an element $\epsilon$ such that $0<|\epsilon|<1/n$ for all $n$). Other authors assume some saturation in the definition of hyperreal number system, see the end of this answer for a note on that.
Now a field $R^*$ is a hyperreal number system under the Henle and Kleinberg definition if and only if it is a proper elementary extension of $\mathbb{R}$. Indeed, if $R^*$ is a proper elementary extension, then there is a "standard part" map $\text{st}\colon R^*\to \mathbb{R}\cup\{\pm\infty\}$. Since $\mathbb{R}$ is complete, every element of $R^*$ which is not greater than all elements of $\mathbb{R}$ or less than all elements of $\mathbb{R}$ is infinitesimally close to a unique element of $\mathbb{R}$. That is, if $a\in R^*\setminus \mathbb{R}$, either $a$ is "infinite" ($\text{st}(a) = \pm\infty$), or $a - \text{st}(a)$ is infinitesimal. And the multiplicative inverse of an infinite element is infinitesimal and vice versa. The point is that any proper elementary extension of $\mathbb{R}$ is non-archimedean, with infinite and infinitesimal elements, and hence satisfies the Henle and Kleinberg definition.
Conversely, if $R^*$ contains $\mathbb{R}$ and is elementarily equivalent to it, it is an elementary extension, by model completeness of the theory of real closed fields. And if it contains an infinitesimal, it is a proper elementary extension.
Now we want to characterize the proper elementary extensions of $\mathbb{R}$ as the models of a first-order theory. We add to the language a constant symbol $c_r$ for all $r\in \mathbb{R}$, and let $T = \text{Th}(\langle\mathbb{R},+,\cdot,<,\{c_r\}_{r\in\mathbb{R}}\rangle)$. An elementary extension of $\mathbb{R}$ is the same thing (up to isomorphism) as a model of $T$, so if we want to count hyperreal number systems, we just have to count models of $T$ (remembering to throw away the standard model $\mathbb{R}$).
Of course, just like any first-order theory with infinite models, $T$ has models of all infinite cardinalities greater than or equal to the size of the language, so $T$ has a proper class of models. But we can ask the more refined question "given a cardinal $\kappa$, how many models does $T$ have of size $\kappa$?"
Now it's time to bring out the sledgehammer: $T$ is an unstable complete first-order theory (this means that $T$ defines a linear order on an infinite set of elements in some model - for our $T$, this is obvious, since for any model of $T$, the whole model is linearly ordered by $<$). It's a cornerstone result of Shelah's Classification Theory that if $T$ is unstable and $\kappa$ is an uncountable cardinal greater than or equal to the size of the language, $T$ has $2^\kappa$ many models of size $\kappa$ up to isomorphism (some counting shows that this is the maximal number possible). If you want to get an idea of how things like this are proven, Section 5.3 of Marker's Model Theory contains a version of this argument.
So, for example, there are $2^{2^{\aleph_0}}$-many hyperreal number systems of size $2^{\aleph_0}$.
I want to end by pointing out that it would be reasonable to define a "hyperreal number system" to be not just any proper elementary extension of $\mathbb{R}$, but only one of size $2^{\aleph_0}$ which is $\aleph_1$-saturated, meaning that for any countable set of formulas (with parameters) such that any finite subset is consistent, there is an element of $R^*$ satisfying all of them. You get a structure with these properties when you take an ultrapower of $\mathbb{R}$ by a nonprincipal ultrafilter on a countable set.
See, for example, this MO answer, where it is asserted that the field of real Puiseux series is not a hyperreal field, because it is not $\aleph_1$-saturated.
If you want to count hyperreal number systems in this narrower sense, the answer depends on set theory. If you assume the continuum hypothesis, then any such field is saturated in its own cardinality (since $2^{\aleph_0} = \aleph_1$), and hence there is a unique hyperreal field up to isomorphism! See here for discussion. On the other hand, if CH is false, I believe you still get the maximum number ($2^{2^{\aleph_0}}$) $\aleph_1$-saturated real closed fields of cardinality $2^{\aleph_0}$.
