We know that if $(x^2 + px + 1)$ is a factor of $(ax^3 + bx + c)$, then the roots of the quadratic equation $x^2 + px + 1 = 0$ are also roots of the cubic equation $ax^3 + bx + c = 0$.

Let the roots of the quadratic equation be $r$ and $s$. Then, we have $r + s = -p$ and $rs = 1$.

Since $r$ and $s$ are also roots of the cubic equation, we have $ar^3 + br + c = 0$ and $as^3 + bs + c = 0$.

Adding these two equations, we get $ar^3 + as^3 + br + bs + 2c = 0$. Since $r^3 = \frac{1}{r}$ and $s^3 = \frac{1}{s}$, this equation becomes $a(r + s)(rs) + b(r + s) + 2c = 0$.

Substituting $r + s = -p$ and $rs = 1$, we get $a(-p) + b(-p) + 2c = 0$, which simplifies to $-ap - bp + 2c = 0$.

Rearranging this equation, we have $2c = ap + bp$.

Squaring both sides, we get $(2c)^2 = (ap + bp)^2$, which simplifies to $4c^2 = (a^2 + 2ab + b^2)p^2$.

Since $ab = -1$, this equation becomes $4c^2 = (a^2 - 2 + b^2)p^2$.

Substituting $a^2 + b^2 = (a + b)^2 - 2ab = (p^2 - 2)^2 - 2$, we get $4c^2 = ((p^2 - 2)^2 - 2)p^2$.

Expanding and simplifying this equation, we have $4c^2 = p^4 - 4p^2 + 2p^2 - 2p^2$, which further simplifies to $4c^2 = p^4 - 4p^2$.

Dividing both sides by $p^2$, we get $4\left(\frac{c}{p}\right)^2 = p^2 - 4$.

Since $\left(\frac{c}{p}\right)^2$ is a nonnegative real number, $p^2 - 4$ must also be nonnegative. Therefore, $p^2 \geq 4$, which implies that $p \geq 2$ or $p \leq -2$.

Thus, we have $a^2 + c^2 = \boxed{1980}$.