Find all integers $(x, y)$ such that $14^{x}=y^{2}-2$
If $x \geq 2$, then $14^{x} \equiv 0[4]$, but on the other hand $y^{2}-2 \equiv 2[4]$ if $y$ is even and $y^{2}-2 \equiv 3[4]$ if $y$ is odd. The two sides cannot be equal, so $x<2$

If $x=1$, we have $14=y^{2}-2$, hence the solution $y=4$

If $x=0$, we have $1=y^{2}-2$, or $y^{2}-3=0$

Finally, the only solution is $(1,4)$