Numerical data suggests the following:

Theorem

If:

$$x^p=y^p+z^p$$ $$p \not | xyz$$ $$gcd(x,y,z)=1$$

Then:

$$x^6 \equiv y^6 \equiv z^6 \pmod{p^3}$$

Proof

Let:

$$az \equiv x-py \pmod{p^3}$$ $$by \equiv x-pz \pmod{p^3}$$

$\implies$

$$a^p-(a-1)^p-1 \equiv b^p-(b-1)^p-1 \equiv 0 \pmod{p^3}$$

Data suggests $a^3 \equiv \left(\frac{x}{z}\right)^3 \equiv b^3 \equiv \left(\frac{x}{y}\right)^3 \equiv -1 \pmod{p}$

$\implies$

$$x^6 \equiv y^6 \equiv z^6 \pmod{p^3}$$

Maybe anyone know how to prove it.