Here's a well known construction of an ordered field extension of $\mathbb Q$ that comes from logic and from nonstandard analysis, as mentioned in the answer of @CarlMummert.
Choose a nonprincipal ultrafilter on the natural numbers $\mathbb N$: an ultrafilter is a finitely additive, $0,1$-valued measure defined on all subsets $A \subset \mathbb N$, such that $\mathbb N$ has measure $1$; and $\mu$ is nonprincipal if $\{i\}$ has measure zero for every $i \in \mathbb N$. The existence of a nonprincipal ultrafilter is an exercise in applying the axiom of choice.
Consider the set $\mathbb Q^{\mathbb N}$ which is the set of all sequences of rational numbers. Define an equivalence relation on this set: given $\underline x = (x_i)$ and $\underline y = (y_i) \in \mathbb Q^{\mathbb N}$, define $\underline x \approx \underline y$ if the set of indices $i$ for which $x_i = y_i$ has measure $1$. Let $[\underline x] = [x_1,x_2,x_3,...]$ denote the equivalence class of $\underline x = (x_1,x_2,x_3,...)$. Define addition and multiplication in the obvious way: $[\underline x] + [\underline y] = [\underline z]$ means that the set of indices for which $x_i + y_i = z_i$ has measure $1$, and similarly for multiplication. Define inequality similarly: $[\underline x] < [\underline y]$ if the set of indices for which $x_i < y_i$ has measure $1$. Check that everything is well-defined, and that you get an ordered field.
To embed $\mathbb Q$ into this field, map the rational number $q$ to $[q,q,q,q,q,q,q,q,q,q,...]$. Check that this is an embedding of ordered fields.
To identify a number greater than any natural number, take $[1,2,3,4,5,6,7,8,9,...]$.