Let $ m^2 $ and $ n^2 $ be the two-digit square numbers; we then have $ 4 \leq m, n \leq 9 $. Putting them next to each other yields a number $ 100m^2 + n^2 $, which must be equal to some other square $ x^2 $. Rearranging, we have $ 100m^2 = x^2 - n^2 = (x+n)(x-n) $, so the RHS contains a factor of 100. The biggest possible square is 8181, whose square root is about 90.5, and the smallest is 1616, whose square root is about 40.2, so $ 41 \leq x \leq 90 $. To get the factor of 100, we have two cases: 1. Both $ x+n $ and $ x-n $ must be multiples of 5. In fact, this means $ n = 5 $, $ x $ is a multiple of 5, and $ x-n $, $ x $, and $ x+n $ are consecutive multiples of 5. Trying possibilities up to $ x = 85 $, we see that this case doesn't work. 2. One of $ x+n $ and $ x-n $ is a multiple of 25. Since $ x+n = 25 $ is impossible, the simplest possibilities are $ x-n = 50 $ and $ x + n = 50 $. The case $ x - n = 25 $ implies $ x + n = 4p^2 $ for $ (x+n)(x-n) $ to be a perfect square multiple of 100, and thus $ 57 \leq 4p^2 \leq 77 $ from $ 41 \leq x \leq 90 $. The only possibility is $ 4p^2 = 64 $, which leads to non-integral $ x $ and $ n $. The case $ x + n = 50 $ requires $ x -n = 2p^2 $ for $ (x+n)(x-n) $ to be a perfect square. To have $ x \geq 41 $ we must have $ x - n \geq 32 $, and in fact the lower bound works: $ (50)(32) = 1600 = 40^2 $. Thus $ x = 41 $, and $ x^2 = \boxed{1681} $.