The equation can be written as $x^4 - x^2 + 1 = 21^y$, i.e., $(x^2 - x + 1)(x^2 + x + 1) = 21^y$. The number in the left hand side is an integer, therefore $y$ must be non-negative. It is easy to see that $x^2 - x + 1$ and $x^2 + x + 1$ are co-prime. As $0 < x^2 - x + 1 < x^2 + x + 1$, we can only have $x^2 - x + 1 = 1$, $x^2 + x + 1 = 21^y$ or $x^2 - x + 1 = 3^y$, $x^2 + x + 1 = 7^y$. The first case leads to $x \in \{0, 1\}$ and then to the solution $x = y = 0$. In the second case, from the first equation we have $(2x - 1)^2 = 4 \cdot 3^y - 3$, hence $3 \mid (2x - 1)^2$. It follows that $9 \mid 4 \cdot 3^y - 3$, i.e., $3 \mid 4 \cdot 3^{y-1} - 1$, and therefore $y = 1$. We immediately obtain $x = \pm 2$. Thus, the equation has three solutions: $x = y = 0$ and $x = \pm 2$, $y = 1$.

The equation can be written $(2x^2 + 1)^2 = 4 \times 21^y - 3$. The number in the left hand side is an integer, therefore $y$ must be non-negative. If $y = 0$ we immediately get $x = 0$. For $y = 1$, we get $x = \pm 2$. For $y \ge 2$, the number $4 \cdot 21^y - 3$ is divisible by 3 but not by 9, and so it can not be a perfect square.