Can $a^2+b+2$ and $b^2+4a$ both be perfect squares? Are there any positive integers $a$ and $b$ so that $a^2+b+2$ and $b^2+4a$ are both perfect squares?

No.
$b^2+4a$ is a perfect square $>b^2$, and $b^2+4a \not =(b+1)^2$ (fails by checking parity), so $b^2+4a \geq (b+2)^2$ so $a \geq b+1$
$a^2+b+2$ is a perfect square $>a^2$, so $a^2+b+2 \geq (a+1)^2$ so $b \geq 2a-1$.
Combining, $b \geq 2a-1 \geq 2(b+1)-1=2b+1>b$, a contradiction.
