Consider the number $10^k - 1$, any two numbers that sum to it must have no carries in addition. So if $x + y = 10^k - 1$ with $s(x) = a^2$ and $s(y) = b^2$, then $a^2 + b^2 = s(x)+s(y) = s(10^k - 1) = 9k$. But if $k$ has a $3 \pmod 4$ prime divisor with odd $\nu_p$, then clearly this is impossible, and there are infinitely many such $k$.

Let $S(a)$ be the sum of digits of $a$. We know that $a-S(a)$ is divisible by 9. Since a square can only give residues 0,1,4,7 when divided by 9 it means that every totally square integer is 0,1,4 or 7 mod 9. It is easy to see that you could not make a number that is 3 mod 9 by adding any 2 of these residues therefore every number of the form $9k+3$ works.