Prove that for $\forall m \ge 2$, $\dfrac{1}{2}(a^4 + b^4 + c^4) \mid [(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m}]$. 
Let $(a, b, c)$ be a Pythagorean triple. Prove that for $\forall m \ge 2$, $$\large \dfrac{1}{2}(a^4 + b^4 + c^4) \mid \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m}\right]$$

Let $P_{n} = a^{2^n} + b^{2^n} + c^{2^n}$, we have that $$4(abc)^{2^k}P_{m} + P^2_{m + 1} - P_{m + 2}$$
$$ = 4(abc)^{2^m} \cdot \left(a^{2^m} + b^{2^m} + c^{2^m}\right) + \left(a^{2^{m + 1}} + b^{2^{m + 1}} + c^{2^{m + 1}}\right)^2 - \left(a^{2^{m + 2}} + b^{2^{m + 2}} + c^{2^{m + 2}}\right)$$
$$ = 4 \cdot \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m}\right] + 2 \cdot \left[(ab)^{2^{m + 1}} + (bc)^{2^{m + 1}} + (ca)^{2^{m + 1}} \right]$$
$$ = 2 \cdot \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m} \right]^2$$
So now we need to prove that $\dfrac{1}{2}(a^4 + b^4 + c^4) \mid 2 \cdot \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m} \right]^2$
$\iff \dfrac{1}{2}(a^4 + b^4 + c^4) \mid 4(abc)^{2^k}P_{m} + P^2_{m + 1} - P_{m + 2}$. But I don't know about whether it is sufficient to prove that $\dfrac{1}{2}(a^4 + b^4 + c^4) \mid P_{n}$ for $\forall n \ge 2$.
And why does $(a, b, c)$ have to be a Pythagorean triple?

As you did, this answer uses $P_m$ where
$$P_m=a^{2^m}+b^{2^m}+c^{2^m}$$
We may suppose that $c^2=a^2+b^2$.
Let $$Q_m=(ab)^{2^m}+(bc)^{2^m}+(ca)^{2^m},\qquad R=\frac{a^4+b^4+c^4}{2}$$
Now, let us prove by induction on $m$ that $R\mid P_m$ and $R\mid Q_m$.
We see that $R\mid P_2$ and $R\mid Q_2$ since
$$Q_2=(a^4+a^2b^2+b^4)^2\qquad\text{and}\qquad R=a^4+a^2b^2+b^4$$
Suppose that 
$$R\mid P_m\qquad\text{and}\qquad R\mid Q_m$$
Then, since
$$Q_{m+1}=Q_m^2-2a^{2^m}b^{2^m}c^{2^m}P_m\qquad \text{and}\qquad P_{m+1}=P_m^2-2Q_m$$
we get
$$R\mid P_{m+1}\qquad\text{and}\qquad R\mid Q_{m+1}\qquad\blacksquare$$
