2017/016) Find a,b satisfying $4^a + 4a^2 + 4 = b^2$

The LHS is even hence to put a limit on a we must have
$(2^a+2)^2 <= 4^a + 4a^2 + 4$
or $4^a + 4 * 2^a + 4 <= 4^a + 4a^2 + 4$ or $2^a <= a^2$ or $ a <= 4$
putting a = 0 to 4 one by one we get $a=2,b=6$ and $a=4,b= 18$ are 2 solution