Answer. Two.

Note that no two squares of natural numbers differ by $1$, since $x^2 - y^2 = (x - y)(x + y)$, where the second factor is greater than one. Therefore, the numbers $a(a+2) = (a+1)^2 - 1$ and $b(b+2) = (b+1)^2 - 1$ are not squares. Moreover, the numbers $ab$ and $a(b+2)$ cannot both be squares, otherwise their product $a^2 \cdot b(b+2)$ would also be a square, and then $b(b+2)$ would also be a square. Similarly, among the numbers $(a+2)b$ and $(a+2)(b+2)$, at most one can be a square. Thus, there are at most two squares among the six numbers.

Two squares can be obtained, for example, for $a = 2$ and $b = 16$: then $a(b+2) = 6^2$ and $(a+2)b = 8^2$.

Remark. There are other examples, for instance, $(a, b) = (6, 96)$.