Prove that formula is satisfiable in some infinite structure, but invalid in all finite structures

The idea is the following : $P$ is interpreted as a relation that is transitive and antisymmetric, in other words, a strict ordering $<$.

But the first part of the formula in question also says the following : there is no maximal element, in other words, for every element, there is a strictly bigger one.

If you have such an interpretation, pick an element $x_0$. Then it's not maximal, so there must be $x_1>x_0$. Similarly, there must be $x_2>x_1$. Proceed on, with induction to create $x_n$ for all $n\in \mathbb{N}$. If $n<m$, then $x_< x_{n+1}<...< x_m$, and so out of transitivity, $x_n < x_m$, and so if $n\neq m$, $x_n\neq x_m$. Now can you have a finite interpretation and such a sequence in it ?

I like to think about this visually/graphically. So think about any $P(x,y)$ relation as there being an arrow from $x$ to $y$.

Because of the relation being asymmetical and transitive, you can never have a cycle of arrows, and therefore we are able to arrange all objects in the domain such that every arrow will go from left to right.

But if the domain is finite, then there must be one or more right-most objects, meaning that they don't have an outgoing arrow ... but that contradicts the first conjunct that requires every object to have an outgoing arrow. So, it is impossible to satisfy this with a finite domain.

If $\Bbb M$ is any interpretation let $P^M=\{(x,y)\in M: \Bbb M\text { satisfies } P(x,y)\}$ then $P^M$ is a binary relation on $M.\;$(I.e. $P^M$ is a subset/subclass of $M\times M.$).

Let $\Bbb M$ be a finite interpretation with $n$ members ($n\in \Bbb N$). Suppose $P^M$ satifies the first condition (I.e. the domain of $P^M$ is $M$) and that $P^M$ is transitive. Then $P^M$ cannot be anti-symmetric because $\exists x\;(P^M(x,x)).$

Proof: Take $x_1\in M$ and for $1\leq j\leq n$ let $x_{j+1}\in \{y\in M: P^M(x_j,y)\}.$ There exist $i,j$ with $1\leq i<j\leq n+1$ with $x_i=x_j$ by the Pigeon-Hole Principle. Now transitivity of $P^M$ implies, by induction on $k$, that $P^M(x_i,x_{i+k})$ whenever $1\leq k\leq n+1-i.$ But then we have $P^M(x_i,x_j)$ and $x_i=x_j,$ contradicting anti-symmetry.

Remark: If you prefer, let $M=\{m_i:1\leq i\leq n\},$ and let $x_1=m_1$ and let $x_{j+1}=m_u(j)$ where $u(j)=\min \{v: P^M(x_j,m_v\}$, if you want a recursive algorithm for defining each $x_j$.