Assume $X$ is a scheme over the base-ring $\mathbb Z$ and $t:\text{Spec}\,\mathbb Z \to X$ is an integer shaped point of $X$. Is there necessarily an affine open subscheme $U$ of $X$ through which $t$ factors?
Context: Richard Garner claims on page 5 first paragraph of An embedding theorem for tangent categories that all schemes are microlinear (in the sense of synthetic differential geometry). They claim that I can reduce the proof to the affine case by using that open embeddings of schemes are formally etale. Formally etale means that if I have some $t: \text{Spec } \mathbb Z = \{0\} \to X$, lying in some open affine subspace $U$, then any infinitesimal thickening of $\{0\}$ such as e.g. $D_2(2) = \text{Spec } \mathbb Z[x,y]/(x,y)^3$ lies also in $U$. But to use this I have to know that $t$ lies in some open affine in the first place.
I am equally happy if you can tell me why any scheme is infinitesimally linear (the weaker version of microlinearity).
This answer is somewhat complementary to the discussion in the comments – while they show the statement holds for a reasonable class of schemes, I find a rather ugly one where the statement fails (as Nathan Lowry expected).
Consider the following scheme $X$: it is the union of two copies of $\mathbb{A}^1_{\mathbb{Z}}$, glued together on the open subset corresponding to all the primes except $(2,T)$ and $(3,T)$. Denote as $X_1$ the (isomorphic and open) image of the first affine line in $X$ (with coordonate $T_1$), and $X_2$ the image of the second one (with coordinate $T_2$).
Then take the following morphism $t: \mathrm{Spec}\,\mathbb{Z} \rightarrow X$: on $R_1=\mathbb{Z}[1/2]$, it is the map $R_1 \rightarrow X_1$ given by $T_1\longmapsto 0$. On $R_2=\mathbb{Z}[1/3]$, it is the map $R_2 \rightarrow X_2$ given by $T_2\longmapsto 0$.
(indeed, the two morphisms agree on the spectrum of $\mathbb{Z}[1/6]$).
What happens here is that we have a lack of separatedness – geometrically, distinct morphisms that somehow still agree on the generic fiber.
But I’m not sure that this is the only obstruction, although by Chow’s lemma separatedness (along with finite type, which is rather natural in this case) implies a property which doesn’t look so much weaker than quasi-projectiveness.