The squares modulo $9$ are $\{0, 1, 4, 7\}$. Any integer is congruent modulo $9$ to the sum of its digits (as powers of $10$ are congruent to $1$ modulo $9$). Therefore any totally square integer is congruent modulo $9$ to one of $\{0, 1, 4, 7\}$. Taking all combinations of pairs, the sum of two totally square integers is congruent modulo $9$ to one of $\{0, 1, 2, 4, 5, 7, 8\}$. Positive integers congruent to $3$ (or to $6$), of which there are infinitely many, are not the sum of two totally square integers.

Alternatively, we may let $S(a)$ denote the sum of the digits of a number $a$ and note that $S(a) \equiv 0, 1 \pmod{3}$ when $a$ is totally square. When $n = n_1 + n_2$ is the sum of two totally square integers $n_1, n_2$ and if $n \equiv 0 \pmod{3}$, we have $0 \equiv n \equiv S(n_1) + S(n_2) \pmod{3}$ which implies $S(n_1) \equiv S(n_2) \equiv 0 \pmod{3}$. A square number that is divisible by $3$ actually is divisible by $9$, hence $S(n_1) \equiv S(n_2) \equiv 0 \pmod{9}$ and so $n \equiv S(n_1) + S(n_2) \equiv 0 \pmod{9}$. Therefore, positive integers that are congruent to $3$ or $6$ (mod $9$) cannot be the sum of two totally square integers, e.g. the numbers $3 \cdot 10^k$ for all $k \ge 0$.