No. Every nonprincipal ultrafilter $U$, considered as a partial under $\subseteq$, is a nontrivial product order. To see this, suppose that $U$ is a nonprincipal ultrafilter on $\kappa$.
Partition $\kappa=A\sqcup B$ into two sets with $A\in U$ and $B$ nonempty.
Every $X\in U$ can be written as
$X=(X\cap A)\sqcup (X\cap B)$, and furthermore, $X\subseteq Y$ just
in case $(X\cap A)\subseteq (Y\cap A)$ and $(X\cap B)\subseteq (Y\cap
B)$. Let $P=U\upharpoonright A=\{ X\subseteq A\mid X\in U\}$ and
$Q=P(B)=\{X\mid X\subseteq B\}$. These are both nontrivial and $\langle U,\subseteq\rangle$ is
isomorphic to the product order $\langle
P,\subseteq\rangle\times\langle Q,\subseteq\rangle$ by the map $X\mapsto (X\cap A,X\cap B)$.

Indeed, you don't even need the ultrafilter to be non-principal, provided $\kappa\geq 3$. The reason is that if $\kappa\geq 3$, then you can partition $\kappa=A\sqcup B$ where $A\in U$, $B$ is nonempty and $A$ has at least two points. In this case, both $P$ and $Q$ again will have at least two elements each, and the rest of the argument is as before.

(Meanwhile, if $\kappa=1$ or $\kappa=2$, then $U$ has only one or two elements, respectively, and so it is not a nontrivial product.)