<s>[INST] Is there an infinite number of numbers like $1600$? My reputation is at this moment at $1600$.
I did some experimenting with $1600$ and obtained the following:
Evidently, it is a perfect square $1600=40^2$
Also, it is a hypothenuse of a Pythagorean integer-triple triangle $1600=40^2=32^2+24^2$.
Also, it can be written as the sum of four non-zero squares $1600=20^2+20^2+20^2+20^2$
So, $1600$ is a perfect square, a hypothenuse of a Pythagorean integer-triple (so can be written as a sum of two non-zero squares), and a sum of four non-zero squares.

Is there an infinite number of numbers like $1600$? Can you find some more?

Edit 1: It is also a sum of $4$ non-zero positive cubes:  $1600=8^3+8^3+8^3+4^3$
Edit 2: It is also a sum of powers from $1$ to $4$, as we see $1600=7^1+9^2+6^3+6^4$
 [/INST] Yes, there are an infinite number of them.  Given your result for $1600$ we can say that $1600k^2$ is another one.  It is a square, it is $(32k)^2+(24k)^2$ and it is $4\cdot (20k)^2$.  We can use the parameterization of Pythagorean triples to get others.  If you choose $m,n$ relatively prime and of opposite parity, they generate a primitive Pythagorean triple $m^2-n^2, 2mn, m^2+n^2$, so if we choose $m,n$ as legs of a Pythagorean triangle the hypotenuse of that triangle will be a square.  As an example, let $m=4,n=3$, which gives the triangle $7,24,25$.  The number $25$ is a square, the sum of two squares, and the sum of four squares as $16+4+4+1$.  Again you can multiply it by any square.  
With the edit, we can still say that $1600k^6$ is a solution by the same reasoning as above plus $1600k^6=3\cdot(8k^2)^3+(4k^2)^3$.
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