This question is answered in "Introduction to Number
Theory" by Niven, Zuckerman& Montmogery (pp.318-319 of
the fifth edition). I summarize their proof below.
Every integer $\geq 34$ is a sum of five positive squares
(while $33$ is not). The number five is optimal, because
the only representation of $2^{2r+1}$ as a sum of four
squares is $0^2+0^2+(2^r)^2+(2^r)^2$ (easy exercice by induction
on $r$).
One can check by hand by noting all the numbers between $34$ and $169$
are sums of five positive squares. Now, let $n\geq 169$ and let us
show that $n$ is a sum of five positive squares.
We know that $n-169$ is a sum of four not necessarily positive
squares, $n-169=x_1^2+x_2^2+x_3^2+x_4^2$ and we can assume
$x_1 \leq x_2 \leq x_3 \leq x_4$.
If $x_1>0$, writing $n=13^2+x_1^2+x_2^2+x_3^2+x_4^2$ we are done. So assume
$x_1=0$.
If $x_2>0$, writing $n=5^2+12^2+x_2^2+x_3^2+x_4^2$ we are done. So assume
$x_2=0$.
If $x_3>0$, writing $n=3^2+4^2+12^2+x_3^2+x_4^2$ we are done. So assume
$x_3=0$.
If $x_4>0$, writing $n=2^2+4^2+7^2+12^2+x_4^2$ we are done. So assume
$x_4=0$.
So now all the $x_i$ are zero, and $n=169=5^2+6^2+6^2+6^2+6^2$. This concludes
the proof.
Try
$$ 31 \cdot 16^n $$
It begins with the observation that $x^4 \equiv 0, 1 \pmod {16}$ but there is more to be done
For $31 \cdot 16^n$ with $n \geq 1,$ if any of the fourth powers are odd, there are at least sixteen odd fourth powers. On the other hand, if we have all even numbers, then dividing each by $2$ gives a representation of $31 \cdot 16^{n-1},$ so induction says sixteen are required
Best Answer
Assume there are $4$ such non-zero squares that add up to $2^{2n+1}$ for any $n \ge 1$, i.e., you have
$$2^{2n+1} = a^2 + b^2 + c^2 + d^2 \tag{1}\label{eq1A}$$
However, note all perfect squares are congruent to $0$, $1$ or $4$ modulo $8$. Since you just asked for a hint, the rest of the answer is in the spoiler below.