(Olimpíada Matemáticos por Diversão 2020)

Find all pairs of integers $(x, y)$ such that:

$$
x^{4}-6 x^{2}+1=7 \cdot 2^{y}
$$
If $y=0$ or $y=1$, we have $x^{4}-6 x^{2}=6$ or 13, so $x^{2}$ divides 6 or 13 (which have no divisors being perfect squares), so $x=1$ but this is not a solution.

If $y=2$, then $x^{2}$ divides 27, so $x=1$ or $x=3$, and we find the solution $(3,2)$.

If $y>2$, then $x^{4}-6 x^{2}+1 \equiv x^{4}+2 x^{2}+1 \equiv\left(x^{2}+1\right)^{2} \equiv 0(\bmod 8)$. However, this implies that $x^{2}+1$ is divisible by 4, which is impossible since 3 is not a possible value for $x^{2}$ modulo 4.