There are two questions:


1. How to prove that in general 


$[\hat{A}(\mathbb HP^m)]\_{4m} = 0$


It is possible to verify it for low values of $m$.
2. How to prove that in general


$\left[\frac{\hat{A}(\mathbb HP^m)} { \hat{M}(\mathbb HP^m) }\right]\_{4m} = 0$


where $\hat{M}(\mathbb HP^m)$ is the Mayer class defined by


$\hat{M}(V) = \prod \_{i=1}^{s}\cosh \left( \frac{y\_{{i}}}{2} \right)$


with


$p(V) =\prod \_{i=1}^{s}(1+{y\_{i}}^2)$.


It is possible to verify it for low values of $m$.



Is true that $\left[\frac{\hat{A}(\mathbb HP^m)} { \hat{M}(\mathbb HP^m) }\right]_{4m} = 0$?
Since $HP^n$ is a spin manifold and since the homogeneous metric has positive scalar curvature, the $\hat{A}$-genus of $HP^n$ is zero by the index theorem and the Weitzenboeck-Lichenrowicz formula.


I cannot at the moment answer the question for the quotient by the Mayer class. But there is a direct computation for the $\hat{A}$-genus, which might help for the quotient as well. I learnt this method from Hirzebruch, when he was still around at the Max-Planck-Insitute.


In my lecture notes <http://wwwmath.uni-muenster.de/u/jeber_02/skripten/mainfile.pdf>, page 161, I discuss the computation of the basic characteristic classes of $HP^n$.


Consider the circle bundle $q: CP^{2n+1} \to HP^n$. $q$ induces an injection on cohomology. Therefore, it is enough to calculate that $(q^{\ast}\hat{A}(THP^n),[CP^{2n}])=0$.
One can show that $q^{\ast} THP^n \oplus H^{\otimes 2} = TCP^{2n+1}$, where $H$ is the Hopf bundle. Denote by $x \in H^2 (CP^{2n+1})$ the generator (first Chern class of $H$).


For a general even power series $F$ with associated multiplicative sequence, we therefore have $q^{\ast} F(THP^n) = F(x)^{2n+2}F(2x)^{-1}$. We have to determine the $2n$th coefficient and prove that it is zero for $n >0$, when $F$ is the power series of the $\hat{A}$-genus. The $2n$th coefficient is $\frac{1}{2 \pi i}\int F(x)^{2n+2}F(2x)^{-1} x^{-2n-1} dx$, integration is over a circle around $0$ in the complex plane.
Form the generating series 
$$
G(t):=\sum\_{n \geq 0} \langle F(x)^{2n+2}F(2x)^{-1}; [CP^{2n}] \rangle t^{2n}
$$
Using the above formula for the coefficients and the sum formula for the geometric series, one sees that the generating series is the same as 
$$
\frac{1}{2 \pi i}\int \frac{F(x)^2}{F(2x)x} \frac{1}{1- \frac{F(x)t}{x}} dx.
$$
For the $\hat{A}$-genus, $F(x) = \frac{x/2}{\sinh(x/2)}$. Perform the substitution $\sinh(x/2)=u$ in the above integral. Finally, you arrive at the integral
$$
\frac{1}{2 \pi i}\int \frac{1}{u-t/2} du.
$$
The value is independent of $t$, as long as $|t|$ is small enough. Therefore $G(t) \equiv 1$, and this proves the result.
I have not performed the first sanity check for this calculation (plug in the Hirzebruch $L$-class has to give the correct value for the signature of $HP^n$, namely $1$). 


