Represent there are infinitely many in FOL
The existing answers provide examples of contexts where "there are infinitely many" can be expressed. However, there is an important sense in which "there are infinitely many" cannot be expressed in a first-order way without some additional context restrictions:

Specifically, fix your favorite first-order language $\Sigma$. Then it's a consequence of the compactness theorem that there is no $\Sigma$-sentence $\varphi$ which is true in exactly the infinite $\Sigma$-structures:


If there were, then $\neg\varphi$ would hold in exactly the finite $\Sigma$-structures.
Then the $\Sigma$-theory $\{\exists x_1...x_n(\bigwedge_{1\le i&lt;j\le n}x_i\not=x_j): n\in\mathbb{N}\}\cup\{\neg\varphi\}$ would be consistent (since any finite fragment holds of any sufficiently large $\Sigma$-structure), so by compactness has a (presumably non-unique) model $\mathcal{M}$.
But $\mathcal{M}$ must be infinite, since any structure satisfying $\exists x_1...x_n(\bigwedge_{1\le i&lt;j\le n}x_i\not=x_j)$ has at least $n$ many elements. By assumption on $\varphi$, this means $\mathcal{M}\not\models\neg\varphi$, and we have a contradiction.


Phrased most relevantly to your question, what we've shown is that even the sentence "There are infinitely many $x$ satisfying $x=x$" is not expressible in first-order logic.



On the other hand, when we restrict attention to specific structures we often have a better situation:


There are many natural structures over which the quantifier "there are infinitely many $x$ such that" is expressible. For example, in $(\mathbb{N}; +,\times,&lt;)$, we may write "there are infinitely many $z$ such that $\varphi(z)$" as "$\forall x\exists y(x&lt;y\wedge \varphi(y))$."


The other two answers fall into this category: they exhibit structures over which "there are infinitely many" is a first-order definable generalized quantifier.