= Solution
Read the initial universal <quantifiers> as governing the conjunction of irreflexivity and transitivity. Explicitly, the intended <first-order sentence> asserts
$$
\forall x\forall y\forall z\bigl(\neg Q(x,x)\land
((Q(x,y)\land Q(y,z))\to Q(x,z))\bigr)
\quad\land\quad\forall x\exists y\,Q(x,y).
$$
Thus $Q$ is an <irreflexive relation>, a <transitive relation>, and a <serial relation>.
Suppose a <first-order model> had a finite nonempty domain of <cardinality> $N$. Choose $a_0$. Seriality supplies successive elements $a_1,\ldots,a_N$ with $Q(a_i,a_{i+1})$. By the <pigeonhole principle>, $a_i=a_j$ for some $i<j$. Repeated transitivity along that part of the chain gives $Q(a_i,a_j)=Q(a_i,a_i)$, contradicting irreflexivity. Hence \b[there is no finite nonempty model]. This is the <infinite-model property of a serial strict partial order>.
\b[An infinite model exists:] take domain $\mathbb N$ and $Q(x,y)$ to mean $x<y$. It is irreflexive and transitive, and $y=x+1$ witnesses seriality for every $x$.
The quantifier-scope convention matters: giving each printed universal <quantifier> only the immediately following atomic scope would leave the transitivity implication with <free variables> and would not express the claimed theorem. The explicit parenthesisation above records the reading under which the source's requested assertion is valid.
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