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It is true if the integral point $T$ is actually a section (as in your example), because you then blow-up a smooth scheme $X\to {\rm Spec}(\mathbb Z)$ along a smooth center $T\simeq {\rm Spec}(\mathbb Z)$. In general, as $T$ is flat over $\mathbb Z$, the fiber $Y_p$ of $Y$ at a prime $p$ is the blow-up of $X_p$ along $T_p$. At a $p$ ramified for $T\to {\rm Spec}(\mathbb Z)$, $T_p$ is non-reduced and in general $Y_p$ is not smooth.

As an example, take $X=\mathbb P^2={\rm Proj}\ \mathbb Z[x,y,z]$ and $T=V_+(x, y^2-2z^2)$. Then $Y$ has singular fiber at $2$.

 Sorry, I was a little to optimistic on the compatibility of the blowing-up of $X$ with the base change $X_p\to X$. However the conclusion is the same. Suppose for simplicity that the generic fiber of $X$ is geometrically connected. Let $p$ be any prime number and let $(X_p)'$ be the blow-up of $X_p$ along $T_p$. Then we have a canonical closed immersion $(X_p)'\to Y_p$ which commutes with $(X_p)'\to X_p$ and $Y_p\to X_p$. Suppose now that $Y$ is smooth, then as $X_p$ and $Y_p$ are connected and smooth of the same dimension and $(X_p)'\to X_p$ is birational, $(X_p)'\to Y_p$ is an isomorphism. Hence $(X_p)'$ must be smooth too. But in general this is not the case as $T_p$ is not necessarily reduced (in the above example $(X_2)'$ is a normal singular surface).

1

It is true if the integral point $T$ is actually a section (as in your example), because you then blow-up a smooth scheme $X\to {\rm Spec}(\mathbb Z)$ along a smooth center $T\simeq {\rm Spec}(\mathbb Z)$. In general, as $T$ is flat over $\mathbb Z$, the fiber $Y_p$ of $Y$ at a prime $p$ is the blow-up of $X_p$ along $T_p$. At a $p$ ramified for $T\to {\rm Spec}(\mathbb Z)$, $T_p$ is non-reduced and in general $Y_p$ is not smooth.

As an example, take $X=\mathbb P^2={\rm Proj}\ \mathbb Z[x,y,z]$ and $T=V_+(x, y^2-2z^2)$. Then $Y$ has singular fiber at $2$.