With $s = t - 1$ we have $s^3 = t^3 - 3t^2 + 3t - 1$, and the first equation can be written as $x = s^3 + 1$. From the second equation we get $27(s^3)^4 = (x^2 - 1)^3$, hence, by taking cube roots, $3s^4 = x^2 - 1$, i.e. $x^2 = 3s^4 + 1$. Comparing with the square of $x = s^3 + 1$ we obtain
$$3s^4 + 1 = s^6 + 2s^3 + 1, \text{ or}$$
$$s^6 - 3s^4 + 2s^3 = 0, \text{i.e.}$$
$$s^3(s - 1)^2(s + 2) = 0.$$
So, $s = -2, 0, 1$, i.e., $t = s + 1 = -1, 1, 2$ and $x = s^3 + 1 = -7, 1, 2$. The three solutions $(t, x)$ are $(-1, -7)$, $(1, 1)$ and $(2, 2)$.