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!
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 $S^3$.
In dimensions at least 5 you should expect this to be false, because if $P$ is a homology sphere of dimension $n \geq 3$, then $\Sigma^2 P \cong S^{\dim P + 2}$ is naturally triangulated (you may always suspend a triangulation), but it is not a PL structure: the links of the top/bottom vertices are $\Sigma P$, which is not a manifold. Of course, the sphere is PL-able, so this is not a counterexample to your question. But it indicates there's no reason to believe there isn't one - the links of vertices are not automatically spheres for silly reasons anymore.
Now apply Galewski-Stern, which says the following (though note the discussion of the Kirby-Siebenmann invariant is more classical than their paper!).
There is a group $\Theta$, called the homology cobordism group, which fits into a short exact sequence $0 \to \text{ker}(\mu) \to \Theta \to \Bbb Z/2 \to 0$. There is an invariant $\text{ks}(M) \in H^4(M;\Bbb Z/2)$ of topological $n$-manifolds, and if $n \geq 5$ the vanishing $\text{ks}(M) = 0$ is equivalent to $M$ supporting a PL structure. Assuming still that $n \geq 5$ and writing $\beta_{\Theta}: H^4(M;\Bbb Z/2) \to H^5(M;\text{ker}(\mu))$ for the associated Bockstein map, then $\beta_{\Theta} \text{ks}(M)$ vanishes if and only $M$ is triangulable.
The fact that $\beta_\Theta$ is nonzero is essentially Manolescu's 2012 contribution to the triangulation question.
So what you want to find is an example of a $(\geq 5)$-dimensional $M$ so that $\text{ks}(M) \neq 0$ but $\beta_\Theta \text{ks}(M) = 0$. Take $X$ to be the $E8$ manifold, and let $X_k = X \times T^{4 - k}$ be the $k$-dimensional manifold you get by crossing with enough circles. We already know that $\text{ks}(X) \neq 0$, because for any 4-manifold with even intersection form, $\text{ks}(X) = \sigma(X)/8 \pmod 2$, and $\sigma(X) = 8$ in this case. Because Kirby-Siebenmann invariants of products $X \times Y$ are (under the Kunneth decomposition) the same as $\text{ks}(X) + \text{ks}(Y)$, we see that $\text{ks}(X_k) = \text{ks}(X) \neq 0$ for all $k$. But because $\beta_\Theta H^*(X;\Bbb Z/2) = 0$, and $\beta_\Theta$ acts on $H^*(X \times Y;\Bbb Z/2) = H^*(X;\Bbb Z/2) \otimes H^*(Y;\Bbb Z/2)$ via $\beta_\Theta \otimes 1 + 1 \otimes \beta_\Theta$, we see that $\beta_\Theta \text{ks}(X_k) = \beta_\Theta \text{ks}(X) = 0$ for all $k \geq 4$.
Thus by the Galewski-Stern result (and Kirby-Siebenmann's work on PL structures), $X_k$ is triangulable for every $k \geq 5$ but not PL. What is more, because $w_1(X_k) = w_2(X_k)$, these are spinnable, if you so desire.