No, this is not possible.
As Mohammad argued in the comments, such a model $M$ must contain all countable ordinals.
We may assume that $M = L^M$, otherwise $L^M$ 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 $\omega_1^L$. We may assume that $\alpha \lt (\kappa^+)^L$ (i.e. $L \vDash |M| = \kappa$). Otherwise we could apply Löwenheim-Skolem in $L$ to get $N \prec M$ with $\kappa \subseteq N$ and $|N| = \kappa$; the transitive collapse of $N$ would then be an uncountable model of ZFC different from $M$.
There are now two cases:
If $\alpha \gt \kappa$, then the poset $(2^{\lt\kappa})^L$ is in $M$. Working in $L$, using the fact that $2^{\lt\kappa}$ is $(\lt\kappa)$-closed and that $|M| \leq \kappa$, we can easily construct a $M$-generic $G$ for $2^{\lt\kappa}$. Then, the generic extension $M[G]$ is an uncountable transitive model of ZFC different from $M$.
If $\alpha = \kappa$ then 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.