All smooth manifolds are triangulable, as you say. This follows from Morse theory, which dictates that you only need to know how to triangulate (PL) handle-attachments, which one can do by hand. The result, though not phrased then in terms of Morse theory, has been known since the 30s. So in my post I meant "topological manifold" when I said "non-triangulable manifold", because that is the only thing that could have made sense!
For dimensions $D \leq 4$ any triangulable manifold is smooth with no further requirement.
For any dimension $D \geq 5$ there are spin manifolds which are triangulable, but do not admit any smooth structure.
So the answer to your title question is $D \leq 4$, and no reasonable assumption on $M$ for $D \geq 5$ is likely to give you a similar result in higher dimensions. Certainly spin is not enough.
As for "triangulable iff smooth". The standard statement is that the notion of PL and smooth structures are equivalent in dimensions up to 6 (in dimension 7, every PL structure may be smoothed, but not uniquely). See here for a good discussion.
A PL manifold can be thought of as a topological manifold $X$ equipped with a(n equivalence class of) triangulation so that the link of any vertex is PL-homeomorphic to $S^{\dim X - 1}$.
In dimension 4 it is already difficult to prove that "triangulable and PL-able" are equivalent. In any dimension, it is not hard to show that $\text{lk}(\sigma)$ is a homology manifold using the cone-homeomorphism from $\text{lk}(\sigma) \times \Bbb R$ to an open subset of a topological manifold; in particular, $\text{lk}(\sigma) \times \Bbb R$ is a topological manifold. It is similarly not hard to show that $\text{lk}(\sigma)$ must have the same homotopy type as $S^{\dim M - 1}$. In dimensions $\leq 2$, homology manifolds are in fact manifolds (Theorem 16.32 in Bredon's sheaf theory) and topological manifolds admit unique PL structures, so we see that there is a PL homeomorphism $\text{lk}(\sigma) \cong S^{\dim M - 1}$. In particular, any triangulation of a manifold of dimension at most $3$ equipped is automatically PL.
In dimension 4 the same is true but one needs to do more work: now you need to use the fact that $\text{lk}(\sigma)$ is itself a simplicial complex. A theorem of Edwards (theorem 3.5 here) then indicates that $\text{lk}(\sigma)$ is a manifold, and hence by the Poincare conjecture that it is homeomorphic to $S^3$.
What you want now is an example, in each dimension $n \geq 5$, of a triangulable manifold $X_n$ which is not PL (and hence not smooth). This is discussed on MO here, citing Rudyak: if $X_4$ is Freedman's E8 manifold, then $X_{4+k} = X_4 \times T^k$ is triangulable, but not PL. This is proved by a sort of dimensional reduction for both parts: in Rudyak's Theorem 7.4, he argues that none possess a PL structure by passing to the universal cover $\widetilde{X}_{4+k} = X_4 \times \Bbb R^k$. The Kirby-Siebenmann product theorem (relating PL structure sets on $M$ and $M \times \Bbb R$) states that this carries a PL structure for any $k \geq 1$ if and only if $\widetilde X_5$ does. Because PL 5-manifolds carry smooth structures, $\widetilde X_5$ is smoothable; one then argues by constructing a bordism between $X_4$ and a smooth spin manifold with the same signature, which is impossible by Rokhlin's theorem. Therefore no $\widetilde X_n$ is PL for $n \geq 4$, and hence neither is any $X_n$ for $n \geq 4$.
To see that $X_n$ is triangulable for $n \geq 5$, see Rudyak's Theorem 21.5: that every compact orientable 5-manifold is triangulable is a theorem of Siebenmann. Then $X_{5 + k} = X_5 \times T^k$ is also triangulable, being a product of triangulable manifolds.
Here is a reasonable approach one might try to see that any orientable 5-manifold is triangulable, but I think it is circular, or at least circuitous.
That any orientable 5-manifold is triangulable might also follow (but see below) from Galewski-Stern. One of the relevant theorems is that if $$0 \to \text{ker}(\mu) \to \Theta \xrightarrow{\mu} \Bbb Z/2 \to 0$$ is the short exact sequence with $\Theta$ the 3-dimensional homology cobordism group and $\mu$ the Rokhlin homomorphism (take a spin manifold $X$ that a homology 3-sphere $\Sigma$ bounds; then $\mu([\Sigma]) = \sigma(X)/8 \mod 2$), then if $\Delta(M) \in \Bbb H^4(M; \Bbb Z/2)$ is the Kirby-Siebenmann class and $\beta_\Theta: H^*(M;\Bbb Z/2) \to H^{*+1}(M; \text{ker}(\mu))$ the associated Bockstein map, then a closed manifold $M$ of dimension $n \geq 5$ is triangulable if and only if $\beta_\Theta \Delta(M) = 0 \in H^5(M;\text{ker}(\mu))$.
(Manolescu's [2013 contribution] was essentially that $\beta_\Theta$ is not identically zero, which is equivalent to the group theoretic statement that $\mu: \Theta \to \Bbb Z/2$ has no section.)
Any short exact sequence $0 \to H \to G \to K \to 0$ gives rise to a long exact sequence on cohomology (with boundary map the Bockstein). If $M$ is an oriented closed manifold of dimension $n$, then the end of this sequence is precisely $$H^{n-1}(M;K) \xrightarrow{\beta} H^n(M; H) \to H^n(M; G) \to H^n(M; K) \to 0;$$ because $H^n(M; A) \cong A$ naturally for an oriented closed $n$-manifold, the end of this sequence is precisely our original short exact sequence $H \to G \to K \to 0$; by exactness, we see that $\beta: H^{n-1}(M; K) \to H^n(M; H)$ is identically zero.
In particular, if $M$ is a closed oriented 5-manifold, we must have $\beta_\Theta \Delta(M) = 0 \in H^5(M; \text{ker} (\mu))$. Therefore, $M$ is triangulable.
But... the Galewski-Stern paper relies on the Siebenmann paper in which compact oriented 5-manifolds are shown more quickly to be triangulable.
