By definition, $T$ is sufficient if $$P(X_1=x_1,X_2=x_2|T=t) \text{ doesn´t depend on }\mu.$$
I think that $T$ isn't sufficient, so I think that if I calculate the probability $$P(X_1=x_1,X_2=x_2|T=0) \text{, it will depend on }\mu.$$
\begin{eqnarray*} P(X_1=x_1,X_2=x_2|T=0)&=& \frac{P(X_1=x_1,X_2=x_2,T=0)}{P(T=0)}\\ &=& \frac{P(X_1=x_1,X_2=x_2,X_1=X_2)}{P(X_1=X_2)}\\ &=&\frac{P(X_1=x_1,X_2=x_2)\cdot 1_{\{x_1=x_2\}}(x_1,x_2)}{P(X_1=X_2)}\\ \text{ by independence }\\ &=& \frac{\frac{\mu^{x_1}e^{-\mu}}{x_2!}\cdot \frac{\mu^{x_2}e^{-\mu}}{x_!}\cdot 1_{\{x_1=x_2\}}(x_1,x_2)}{P(X_1=X_2)}. \end{eqnarray*}
It's possible to simplify this probability?
There's no need to keep separate values $x_1$ and $x_2$: If they differ, the probability is zero, which is obviously independent of $\mu$; so only the case $X_1=X_2=x$ is of interest. In that case the numerator is
$$ \frac{\mu^{2x}\mathrm e^{-2\mu}}{x!^2} $$
and the denominator is
$$ \sum_{k=0}^\infty\frac{\mu^{2k}\mathrm e^{-2\mu}}{k!^2}=\mathrm e^{-2\mu}I_0(2\mu)\;, $$
where $I_0$ is a modified Bessel function of the first kind. Thus
$$ P(X_1=X_2=x\mid T=0)=\frac{\mu^{2x}}{x!^2I_0(2\mu)}\;, $$
which depends on $\mu$. So you were right.