Actually, there is a (nontrivial) theorem somewhere in the vicinity of what you are trying to ask. The right objects to consider are triangulated manifolds (more precisely, PL triangulations, of manifolds, i.e. where links are PL spheres). (PL stands for "piecewise-linear," this is a generalization of the notion of a piecewise-linear function you might be familiar with.) Every triangulated manifold $M$ defines a graph (the 1-dimensional skeleton of the triangulation), but a triangulation actually contains much more information than that graph.
Every manifold admits infinitely many triangulations, provided that it admits one. Thus, the natural notion replacing diffeomorphism for smooth manifolds is the one of a PL homeomorphism. Equivalently, you can say that two triangulations are regarded as "the same" if they admit isomorphic subdivisions. Thus, one defines the notion of PL isomorphic triangulated manifolds. Now, one can ask:
Is there a (PL) triangulated manifold which is homeomorphic to ${\mathbb R}^4$ (equipped with the "standard" PL structure, obtained by a suitable subdivision of the standard cubulation) as a topological space but is not PL isomorphic to such ${\mathbb R}^4$.
The answer to this is indeed positive and the proof uses the result about the existence of exotic differentiable structures on ${\mathbb R}^4$. The proof boils down to nontrivial a theorem (due to Kirby and Siebenmann) that in dimensions $\le 6$ the categories PL and DIFF are naturally isomorphic. In particular, if $M, M'$ are smooth 4-dimensional manifolds which are homeomorphic but not diffeomorphic then they can be PL triangulated so that the resulting PL manifolds are not PL isomorphic. From this, it follows that for every exotic smooth ${\mathbb R}^4$ there exists an exotic PL triangulated ${\mathbb R}^4$.
In dimension 4, one has the following equivalence:
The set $\mathcal B_4$ of compact smooth manifolds homeomorphic to $B^4$, considered up to oriented diffeomorphism, is in canonical bijection with the set $\mathcal S_4$ of compact smooth manifolds homeomorphic to $S^4$, considered up to oriented diffeomorphism.
Proof. We construct maps both ways, which will be clearly inverse.
$C: \mathcal B_4 \to \mathcal S_4$. Pick $B \in \mathcal B_4$. By the resolution of the 3D Poincare conjecture, there is some oriented diffeomorphism $\varphi: \partial B \to S^3$. One may then "cap off the boundary": define $$C(B) = B \cup_\varphi B^4,$$ defined as $B \sqcup B^4$, modulo identifying $\partial B \cong S^3$ via the diffeomorphism $\varphi$.
$C$ is well-defined by Cerf's theorem that $\pi_0 \text{Diff}^+(S^3) = 1$: all diffeomorphisms are the same up to isotopy. Isotoping $\varphi$ above does not change the diffeomorphism type; so $C$ is a set map.
Conversely one has $D: \mathcal S_4 \to \mathcal B_4$, with $D(S)$ given by deleting the interior of some oriented embedding $\iota: B^4 \to S$. Oriented embeddings of balls into a connected manifold are unique up to isotopy (Palais, but straightforward); this is true in all dimensions.
Clearly $D(C(B)) = B$ (delete the ball you glued) and $C(D(S)) = S$ (glue in the ball you deleted). Therefore $C,D$ are inverse bijections.
However one has the following for $n \geq 6$.
The set $\mathcal B_n$ is trivial.
Proof: Pick $B \in \mathcal B_n$. Delete a standard ball from its interior; then we are provided with a compact manifold $W$, with $\partial W = S^{n-1} \sqcup \partial B$, so that $W \cup_{S^{n-1}} B^n = B$.
$W$ is an h-cobordism (algebra). Therefore by the h-cobordism theorem there is a diffeomorphism $W \cong S^{n-1} \times [0,1]$, which sends $S^{n-1}$ to $S^{n-1} \times \{0\}$ by the identity map. Therefore $W \cup_{S^{n-1}} B^n \cong B^n$. Therefore $B \cong B^n$.
Of course there are no exotic n-balls n = 1,2,3. For n=5 I do not know. I think the answer is that $\partial: \mathcal B_5 \to \mathcal S_4$ is an injection (maybe bijection?) but I don't know off the top of my head. Some googling or looking on Math Overflow should help. Obviously "exotic balls" will get you nowhere but some buzzwords like diffeomorphism might.
Best Answer
Yes, it is true. For $n>8$ there exist exotic $n$-spheres if $n$ is congruent modulo 192 to one of 2, 6, 8, 10, 14, 18, 20, 22, 26, 28, 32, 34, 40, 42, 46, 50, 52, 54, 58, 60, 66, 68, 70, 74, 80, 82, 90, 98, 100, 102, 104, 106, 110, 114, 116, 118, 122, 124, 128, 130, 136, 138, 142, 146, 148, 150, 154, 156, 162, 164, 170, 178, 186.
See Corollary 1.5 of "The 2-primary Hurewicz image of tmf" by Behrens, Mahowald, Quigley.