$a^b = 1024$

We know that $1024 = 2^ {10} = 2^{2\times 5}$ , hence either $a=2$ and $b=10$ or $a=2^2 =4$ and $b=5$ or $a=2^5=32$ and $b=2$ .

Don't forget about negative integers : $(-2)^{10}= 2^{10}$ and $(-32)^2 = 32^2$ .

So in total $5$ couples of integers $(a,b)$ satisfy the equation : $\{(-32,2),(-2,10),(2,10),(4,5),(32,2)\}$ .