Let Josh's age be $10a+b$, and let Fred's be $10b+a$. The difference between the squares is $99(a^2-b^2)$. Then clearly we must have that $a^2-b^2 = (a-b)(a+b)$ be of the form $11k^2$. Since $0 \leq a, b \leq 9$, we have that $a+b \leq 18$ and thus we must have that $a+b=11$, $a-b=k^2$. This only occurs when $a=6,b=5$, and indeed we get $6+5=11, 6-5=1^2$. So we have the ages of Josh and Fred as $65, 56$. Checking, we get that $65^2-56^2 = 33^2$, so we're done.