Actually, I am interested in a non-Noetherian ring in which $\mathrm{Z}$ can be elementary embedded. The first candite that came to my mind was $\mathrm{Z} [x_1, x_2, \dots]$.
I was trying to use the Tarski-Vaught test, but it was unsuccessful.
Is it true? Or is my assumption wrong?
P.S. I consider $\mathrm{Z}$ in the language $\{ =, 0, 1, +, \cdot \}$.
$\Bbb Z$ satisfies the formula "for all $n$, either $n$ or $-n$ is the sum of four squares" (by this theorem). However $\Bbb Z[x_1, x_2, \dotsc]$ does not satisfy this - for example this can't be the case for $x_1$ because this polynomial takes on both negative and positive values. So they aren't even elementarily equivalent!
(Added: an even easier way to distinguish is that in $\Bbb Z$, every element is of the form $2k$ or $2k + 1$ for some $k$, but this is not true in the polynomial ring - consider $x_1$ again. This argument shows they aren't even elementarily equivalent as additive groups. See also this)
To find a non-Noetherian elementary extension of $\Bbb Z$, consider the elementary diagram of $\Bbb Z$ (just as in the proof of upward Löwenheim–Skolem, this gives a theory each of whose models has $\Bbb Z$ as an elementary substructure when viewed as rings).
Now add constants $c_1, c_2, \dotsc$, and sentences "$c_i$ is not in the ideal generated by $c_1, \dotsc, c_{i - 1}$" for each $i$.
Observe that $\Bbb Z$ has arbitrarily long chains of ideals, so is a model of any finite fragment of this new theory. So by compactness, this new theory has a model, which has $\Bbb Z$ as an elementary substructure, and which has an infinite ascending chain of ideals.
The above technique works just as well to show that any ring with arbitrarily long chains of ideals has a non-Noetherian elementary extension. This condition is actually necessary, since if chains of ideals all have length at most $N$, then the first-order sentence "for all $c_1, \dotsc, c_N$, if $c_k$ is not in the ideal generated by $c_1, \dotsc, c_{k - 1}$ for each $k$, then the ideal generated by the $c_i$ is the entire ring" is part of the complete theory of the ring you started with, and no ring with this property can be non-Noetherian. (possibly replace $N$ with $N \pm 1$ depending on how you interpret "chain of ideals" and "length")
In this special case where we're looking at $\Bbb Z$, we can actually say a bit more, though. The following discussion will require a bit more technical background than the first part of my answer, but I thought it was interesting, so I'd mention it. As Tomasz' comment alludes to, there are in fact no Noetherian proper elementary extensions of $\Bbb Z$. A closely related question is this one. Essentially, the key observation is that by the four-square theorem, the naturals are definable in $\Bbb Z$. Therefore all of the machinery that has been developed to understand models of Peano Arithmetic can be applied to this problem. (In fact there's a one-to-one correspondence between isomorphism types of elementary extensions of $\Bbb Z$ and isomorphism types of elementary extensions of $\Bbb N$, given by "restricting to the positive part" in one direction, and "adding in the negative numbers" in the other direction).
This tells us firstly that we can explicitly find an ascending chain of ideals. The function $\Bbb N \to \Bbb N$ given by $n \mapsto 2^n$ is definable in $\Bbb Z$, by standard $\mathsf{PA}$-trickery involving for instance the $\beta$-function, and it's a theorem that for all $n$, $2^n$ is a proper divisor of $2^{n + 1}$.
If $M$ is some nonstandard elementary extension of $\Bbb Z$, let $n$ be a nonstandard positive element. Then the chain given by $(2^n) \subseteq (2^{n - 1}) \subseteq (2^{n - 2}) \subseteq \dotsb$ is an infinite ascending chain of properly nested ideals. So we are done!
Secondly, this tells us that it is quite hard to write down any nontrivial elementary extension of $\Bbb Z$, particularly using nice algebraic tools! By the above, such an extension contains a nonstandard model of $\mathsf{PA}$, which are known to be non-computable. So $\Bbb Z[x_1, x_2, \dotsc]$ really never stood a chance.
The ultrapower construction does give a somewhat explicit way to get your hands on such an extension (in general you can always re-phrase a compactness proof as an ultraproduct proof).