This is Problem 20-A in Milnor-Stasheff's book Characteristic Classes:
Let $\tau$ be the tangent bundle of $\mathbb{P}^m(H)$, where $H$ is the quaternion algebra. $\gamma$ is the tautological bundle over $\mathbb{P}^m(H)$. Use the isomorphism $$\tau\oplus\text{Hom}_H(\gamma,\gamma)\cong\text{Hom}_H(\gamma,H^{m+1})$$ to show that $$p(\tau)=\frac{(1+u)^{2m+2}}{1+4u}$$ $u=c_2(\gamma)$ is a generator of $H^4(\mathbb{P}^m(H))$.
I'm not sure what's wrong with my proof: Take Pontrjagin classes on both sides of the isomorphism. Since $p(\text{Hom}_H(\gamma,H))=(1+u)^2$, we have $p(\text{Hom}_H(\gamma,H^{m+1}))=(1+u)^{2m+2}$. Because $\text{Hom}_H(\gamma,\gamma)\cong\epsilon^4$ is trivial bundle, $p(\tau\oplus\text{Hom}_H(\gamma,\gamma))=p(\tau)$. Therefore, $p(\tau)=(1+u)^{2m+2}$.
Where does $1+4u$ come from ? I appreciate any hints or answers.
I think the misconception present in the question is that for a quaternionic vector space $V$, the endomorphism space $\mathrm{Hom}_{\mathbb{H}}(V, V)$ is a quaternionic vector space, just as $\mathrm{Hom}_{K}(V, V)$ would be a $K$-vector space for any vector space $V$ over a field $K$. If this was true, then it would indeed be canonically isomorphic to $\mathbb{H}$, as it has a canonical element, the identity endomorphism.
However, this is not true, due to non-commutativity of the quaternions. Think about how you would define this vector space structure, and you will notice that you only get a vector space structure over the center of $\mathbb{H}$, which is $\mathbb{R}$.
The upshot is that $\mathrm{Hom}_{\mathbb{H}}(\gamma, \gamma)$ is not a rank $1$ quaternionic vector bundle, but a rank $4$ real vector bundle (which actually has a canonical section, given by the identity elements, but this does not really matter).
Let us calculate its first Pontrjagin class.
(1) Recall that $\gamma$ itself is a rank $4$ real vector bundle with three almost complex structures $I$, $J$ and $K$ satisfying the quaternion relations. In particular, it has an associated rank $2$ complex vector bundle $\gamma_{\mathbb{C}}$ using the complex structure $I$.
(2) There is a canonical isomorphism $$\mathrm{Hom}_{\mathbb{H}}(\gamma, \gamma) \otimes_{\mathbb{R}} \mathbb{C} \cong \mathrm{Hom}_{\mathbb{C}}(\gamma_{\mathbb{C}}, \gamma_{\mathbb{C}})$$ given by sending $A \otimes (a + bi) \mapsto a\cdot A + b\cdot AI$. To see that this map is an isomorphism, we show that it is injective (from which the claim follows from dimension reasons): The general element of $\mathrm{Hom}_{\mathbb{H}}(\gamma, \gamma) \otimes_{\mathbb{R}} \mathbb{C}$ has the form $A \otimes 1 + B \otimes i$. Hence if this gets sent to zero under our map, then we would have $A = - BI$. But $A$ commutes with the left $\mathbb{H}$-action, while $BI$ does not, unless $B=0$. But if $B=0$, then also $A=0$.
(3) We can now calculate $$p_1\big(\mathrm{Hom}_{\mathbb{H}}(\gamma, \gamma)\big) = -c_2\big(\mathrm{Hom}_{\mathbb{H}}(\gamma, \gamma) \otimes_{\mathbb{R}}\mathbb{C}\big) = -c_2\big(\mathrm{Hom}_{\mathbb{C}}(\gamma_{\mathbb{C}}, \gamma_{\mathbb{C}})\big) = -c_2(\gamma_{\mathbb{C}} \otimes_{\mathbb{C}} \gamma_{\mathbb{C}}^*).$$
(4) To get a formula for the Chern class of the tensor product, we can use the Chern character. The first terms of the Chern character are $$\mathrm{ch}(E) = \mathrm{rk}(E) + c_1(E) + \frac{1}{2}\big(c_1(E)^2 - c_2(E)\big) + \dots.$$ If now $E$ and $F$ are two vector bundles of rank two with $c_1(E) = c_1(F) = 0$, then the formula $\mathrm{ch}(E \otimes F) = \mathrm{ch}(E)\mathrm{ch}(F)$ yields $$ 4 - \frac{1}{2} c_2(E \oplus F) + \dots = \left(2 - \frac{1}{2} c_2(E) + \dots\right)\left(2 - \frac{1}{2} c_2(F) + \dots\right) = 4 - c_2(E) - c_2(F) + \dots$$ Hence $c_2(E \oplus F) = 2c_2(E) + 2c_2(F)$.
(5) Finishing off the calculation, we get $$ p_1\big(\mathrm{Hom}_{\mathbb{H}}(\gamma, \gamma)\big) = -c_2(\gamma_{\mathbb{C}} \otimes_{\mathbb{C}} \gamma_{\mathbb{C}}^*) = - 2c_2(\gamma_{\mathbb{C}}) - 2c_2(\gamma_{\mathbb{C}}^*) = -4 c_2(\gamma_{\mathbb{C}}) = - 4u,$$ hence $$p\big(\mathrm{Hom}_{\mathbb{H}}(\gamma, \gamma)\big) = 1 - 4u.$$ What's bugging me, though, is the sign difference. Are there different conventions at play here?
Edit:
I get $$p\big(\mathrm{Hom}_{\mathbb{H}}(\gamma, \mathbb{H}) \big) = (1-u)^2,$$ so it seems to me that the generator that Milnor-Stasheff refer to is $u = - c_2(\gamma)$.