P³ – Q³ = 2PQ + 8                                               (1)

Changing to new integer variables that can be separated:

Let        P + Q = a    and   P – Q = b        so that

            PQ  = (a² – b²)/4

and       P² + Q² = (P + Q)² – 2PQ = a² – (a² – b²)/2

(1) becomes  (P – Q)(P² + Q² + PQ) = 2PQ + 8

            b(a² – (a² – b²)/4) = (a² – b²)/2 + 8

                        3a²b + b³ = 2a² – 2b² + 32

Thus     a² = (32 – 2b² – b³)/(3b – 2)                   (2)

The cubic numerator of (2) has positive stationary
values when b = 0 and b = -4/3, and is zero between
b = 2 and b = 3. So it is positive for integer b <= 2.

The denominator of (2) is positive for integer b >=1.

So a is real only when integer b = 1 or 2.

When b = 1, a² = 29, so no integer solution.

When b = 2, a² = 4,   giving a = +2 or –2.

So only two solutions: