Write $n = xb + y$ with digit sum $s = x + y = \sqrt{n}:$ then $s^2 - s = x(b - 1),$ so $b - 1$ must divide $s(s-1)$

Each $s \le b-1$ for which $s(s-1)$ is divisible by $(b-1)$ gives one $n;$ mod $b-1$ there are $2^{\omega}$ such residues, where $\omega$ counts the distinct primes of $b-1$

A two-digit number in base $b$ is $n = xb + y$ with $1 \le x \le b-1$ and $0 \le y \le b-1,$ and the condition says $n = s^2$ where $s = x + y.$ Then $s^2 = xb + y = x(b-1) + s,$ so $$s(s - 1) = x(b - 1).$$ Note $s \le \sqrt{b^2 - 1} \lt b.$ Conversely, for any $s$ with $2 \le s \le b - 1$ and $s(s-1)$ divisible by $(b-1),$ setting $x = \frac{s(s-1)}{b-1}$ and $y = s - x = \frac{s(b-s)}{b-1}$ gives $1 \le x \le b-1$ and $0 \le y \le b-1,$ hence exactly one $b$-eautiful integer $n = s^2.$ So the count equals the number of $s \in \{2, \ldots, b-1\}$ with $s(s-1) \equiv 0 \pmod{b-1}.$

Let $m = b - 1.$ Since $s$ and $s - 1$ are coprime, each prime power dividing $m$ must divide $s$ or $s - 1,$ so by the Chinese remainder theorem there are $2^{\omega(m)}$ solutions modulo $m,$ where $\omega(m)$ is the number of distinct prime factors of $m.$ Among the representatives $1, 2, \ldots, m,$ only $s = 1$ falls outside our range (and $s = m$ qualifies), so the count is $2^{\omega(m)} - 1.$

We need $2^{\omega(m)} - 1 \gt 10,$ i.e. $\omega(m) \ge 4.$ The smallest positive integer with four distinct prime factors is $2 \cdot 3 \cdot 5 \cdot 7 = 210,$ so the least base is $b = 211$ (which has $2^4 - 1 = 15$ $b$-eautiful integers).