The primes below 10 are 4:
So each of your numbers has the form:
$2^a \cdot 3^b \cdot 5^c \cdot 7^d$
There 16 possible tuples (a,b,c,d) when each of $a,b,c,d$ is viewed modulo 2.
We define each pigeonhole as the sequence $(A,B,C,D)$ where A,B,C,D are the residues of a,b,c,d respectively when divided by 2. There are 16 possible pigeonholes ($2 \cdot 2 \cdot 2 \cdot 2$).
This means you can find two numbers $X,Y$ among those 17 such that their tuples
$a_1,b_1,c_1,d_1$ and $a_2,b_2,c_2,d_2$
are such that
$a_1$ and $a_2$ are both odd or both even
$b_1$ and $b_2$ are both odd or both even
$c_1$ and $c_2$ are both odd or both even
$d_1$ and $d_2$ are both odd or both even
Now multiply X and Y and you get a square because
$a_1 + a_2$, $b_1 + b_2$, $c_1 + c_2$, $d_1 + d_2$
will all be even.