No, this is not possible.
As Mohammad argued in the comments, such a model $M$ must contain all countable ordinals.
If $M \neq L^M$ then $L^M$ is another transitive model of ZFC which is uncountable since $\omega_1 \subseteq L^M$.
If $M = L^M$, then $M = L_\alpha$ for some $\alpha \geq \omega_1$. The cardinal $\omega_1$ must be an uncountable regular cardinal in $L$ and I will now write $\kappa$ instead of $\omega_1$ to avoid confusion with $\omega_1^L$.
From now on, work in $L$. We may assume that $\alpha \lt \kappa^+$. Otherwise we could apply Löwenheim-Skolem in $L$ to get $N \prec M$ with $\kappa \subseteq N$ and $|N| = \kappa$; the transitive collapse of $N$ is an uncountable model of ZFC different from $M$.
Suppose $\alpha \gt \kappa.$ The poset $2^{\lt\kappa} = (2^{\lt\kappa})^M$ is $(\lt\kappa)$-closed in $L$. Since $|M| \leq \kappa$, we can construct a $M$-generic $G$ for $2^{\lt\kappa}$ in $L$. Then $M[G]$ is an uncountable transitive model of ZFC different from $M$.
The only remaining case is when $\kappa = \alpha$, in which case every set in $M$ is countable in $V$. Therefore, every set forcing in $M$ has an $M$-generic set in $V$, which leads to a plethora of different uncountable transitive models of ZFC.