Can a Sum of distinct squares ever equal power of two? Does there exist $2^t,\ t\in\mathbb{Z}_+$ which can be express as Sum of two or more distinct square number.

Or
Can it be shown that
$$\begin{split}2^t &\ne \sum a_i^2 = a_1^2+ a_2^2+\cdots+a_n^2\end{split}$$
Where $n\ge 2$ and $\{a_i,t\}\in\mathbb{Z}_+$ and $a_i \ne a_j$ for $1\le i,j \le n$

Example: $2^6=64=7^2+3^2+2^2+1^2+1^2$ here $1^2$ repeat two times so this is not allowed.
My incomplete attempt for arithmetic squares

Edit: check related new post, Can a sum arithmetic square ever equal to power of two?

Let $n,u,d\in\mathbb{Z}_+$
$$\begin{split}\sum_{q=0}^u (n+qd)^2 &=n^2+(n+d)^2+(n+2d)^2+\cdots+(n+ud)^2\\ &=n^2(u+1)+\frac{(u+1)u}{2}(2nd+d^2)+\frac{(u+1)u(u-1)}{3}d^2 \end{split}$$
Let
$$\begin{split}2^t &=\sum_{q=0}^u (n+qd)^2 \\ \implies 3\cdot 2^{t+1}&=6n^2(u+1)+3(u+1)u(2nd-d^2)+(u+1)u(u-1)2d^2 \\ &= (u+1)(6n^2+3u(2nd+d^2)+u(u-1)2d^2)\\ &(in\ case,\ u+1= 3) \\
\implies 2^t&= 3n^2+3(2nd+d^2)+2d^2\\ &= n^2+(n+1)^2+(n+2d)^2 \end{split}$$
Now we need to simplify for case, $6n^2+3u(2nd+d^2)+u(u-1)2d^2=3\cdot2^x$ and $u+1=2^y$ where $x+y=t+1$ but I'm stuck here. Thank you.
Related post:
Can a sum of consecutive $n$th powers ever equal a power of two?
 A: Counterexample: For $t=8, n = 5$ and $(a_1,a_2,a_3,a_4,a_5)=(1,5,7,9,10)$ we have: 
$$2^8=1^2+5^2+7^2+9^2+10^2$$
A: We have, for instance,
$$
169 + 49 + 25 + 9 + 4=256
$$
A: No.  2 is a direct prime in the gaussian integers, which means that some power of it is twice another mumber.  Thus 1+i squared is 2i, whereas 2+i squared is 3+4i.
Since the sum of two squares equate to non-direct numbers squared, it does not happen for direct numbers.
A similar thing is seen in eisenstein numbers, which leads to a^2+ab+b^2 being square.  3 divides instances, but you get free powers only for numbers comprised of primes 6n+1.
A: It would be extremely surprising if the answer were no. Consider just the first $n$ squares. There are $2^n$ ways to construct distinct sums of these, and each sum adds up to something between $0$ and $\frac16(2n^3-3n^2+n)$ which is approximately $\frac{n^3}3$.
But $2^n$ is vastly larger than $\frac{n^3}3$. Consider even a small number like $n=80$. On the one hand we have $2^{80}$ which is the number of atoms in the universe and on the other we have the everyday number 173,880. 
What are the chances that any of the $\frac{n^3}3$ possible sums $S$ is so unusual that none of the $2^n$ possible selections of squares adds up to $S$? There would have to be a very good reason, and it would not be a subtle one.
Instead, we should guess that the opposite is true: except for a few very small exceptions, every power of 2 should be representable as a sum of distinct squares in a great many ways.
