Mark has a proof of vanishing for $\mathrm{Spin}(7)$.
I think it's correct.
Here's a careful statement and an attempt at a proof.

**Theorem:** Suppose $(E,A) \to X$ is a (hermitian) vector bundle with (compatible) instanton connection over a $\mathrm{Spin}(7)$ base (that is, $F^7 = 0$).
There are no harmonic spinors in $\Gamma(S_7^+ \otimes E)$.
That is, if $\psi \in \Gamma(S_7^+ \otimes E)$ and $D\psi = 0$, then $\psi = 0$.

Through the proof it suffices to work in a nice frame $\{e_1, \ldots, e_8\}$ of $TM$ around some point $x$ chosen so that at $x$, Clifford multiplication by the standard basis $\gamma^i$ of $\Lambda^2_7$ is covariant constant.
That is,  at $x$ $$[c(\gamma^i), \nabla] = 0. \tag{1}$$
We define
$$D^7 := \frac{1}{28} \sum_{j=1}^7 \sum_{b=1}^8 c(\gamma^j e_b) c(\gamma^j) \nabla_b \tag{2}$$
$$D^{21} := \frac{1}{56} \sum_{1 \le j < k \le 7} \sum_{b=1}^8 c(\gamma^{jk} e_b) c(\gamma^{jk}) \nabla_b \tag{3}$$
<WRAP center round info 60%>
Here we are working with the convention that for a skew operator $A$, we have $c(A) = \frac{1}{2} A_{ab} c_a c_b$.
</WRAP>

**Lemma 1:** $D = D^7 + D^{21}$

<WRAP center round todo 60%>
**Proof:** We have
\begin{align*}
D &= c_b \nabla_b \\
&= -\frac{1}{8} c_a c_a c_b \nabla_b \\
&= \frac{1}{8} D - \frac{1}{8} \sum_{a \neq b} c_a c_a c_b \nabla_b
\end{align*}
so
$$D = -\frac{1}{7} \sum_{a \neq b} c_a c_a c_b \nabla_b$$
</WRAP>


**Lemma 2:** If $\psi \in \Gamma(S_7^+ \otimes E)$, there exist 7 sections $\psi_i \in \Gamma(S_1^+ \otimes E)$ such that $\psi = c(\gamma^i) \psi_i$.
Moreover, $c(\gamma^j) c(\gamma^m) \psi = -16 \delta_{jm} \psi$.

**Proof:** Essentially, this is a consequence of the following orthogonality relation: for unit $\psi \in S^+_1$,

Suppose $j \neq k$.
Then
\begin{align*}
\langle c(\gamma^j) \psi, c(\gamma^k)\psi \rangle &= \frac{1}{4} \langle [c(\gamma^{jk}), c(\gamma^k)] \psi, c(\gamma^k) \psi \rangle \\
&= \frac{1}{4}\langle c(\gamma^{jk})c(\gamma^k) \psi, c(\gamma^k) \psi \rangle \\
&= 0
\end{align*}
since, for example, Clifford multiplication is skew-adjoint.
Note we used that $c(\gamma^{jk}) \psi = 0$.
Skew-adjointness gives that $c(\gamma^j) c(\gamma^m) \psi = 0$ if $j \neq m$.

For $j = k$ one requires an eigendecomposition of $\psi$. 
It is a fact that the eigenvalues of $c(\gamma^k)$ acting on $S^+$ are $\pm 4i$ with multiplicity 1 for each sign and 0 with multiplicity 6.
Thus $\psi = \psi_{+} + \psi_{-} + \psi_0$ splits as a sum of eigenvectors for a fixed $k$.
However, $\psi_0 = 0$; 

<WRAP center round todo 60%>
Finish proof that zero part vanishes
</WRAP>
Thus $\psi = \psi_+ + \psi_-$ and $c(\gamma^k)^2 \psi = (4i)^2 \psi_+ + (-4i)^2 \psi_{-} = -16\psi$.

**Lemma 3:** $\ker D = \ker D^7 \cap \ker D^{21}$, the kernels being in $\Gamma(S^+ \otimes E)$.

We have $|D\psi|^2 = |D^7 \psi|^2 + |D^{21}\psi|^2 + 2\langle D^7\psi, D^{21}\psi\rangle$.
It suffices (by the assumption that Clifford multiplication commutes with derivatives) to consider $\psi \in S^+_1$; but then the cross term vanishes a priori
<WRAP center round todo 60%>
do this more carefully later
</WRAP>

**Lemma 4:** If $D^7 \psi = 0$ then $c(\gamma^j e_a) \nabla_a \psi_j = 0$ for all $j$ and $a$.
If $D^{21} \psi = 0$ then $c(\gamma^{jk} e_a) \nabla_a \psi_k = 0$ for all $j, k$ and $a$.

**Proof:** Compute:

\begin{align*}
0 &= \frac{28^2}{16^2}|D^7 \psi|^2 \\
  &= \frac{1}{16^2} \langle c(\gamma^j e_a) c(\gamma^j) \nabla_a c(\gamma^m) \psi_m, c(\gamma^k e_b) c(\gamma^k) \nabla_b c(\gamma^\ell) \psi_\ell \rangle \\
  &= \langle c(\gamma^j e_a) \nabla_a \psi_j, c(\gamma^k e_b)  \nabla_b \psi_k \rangle \\
  &= |c(\gamma^j e_a) \nabla_a \psi_j|^2
\end{align*}

<WRAP center round tip 60%>
We know $\tau_{aj} := \nabla_a \psi_j \in S_1^+$ since $\nabla$ preserves the decomposition.
If it is non-zero, then $c(\gamma^k) \tau_{aj}$ span $S_7^+$ by Lemma 2.
So idea is $c(\gamma^j e_a)$, $c(\gamma^{jk} e_a)$ should span everything and in particular one of them should contain something in $S_7^+$, so this is our contradiction.
</WRAP>