Let $n$ be the number of digits of $b$ and let $b = k a$. Then by the conditions of the problem, $10^n \cdot a + k a = (a + k a)^2$, or
$$a = \frac{10^n + k}{(k+1)^2} \qquad (1)$$
If $k$ were odd, then the numerator on the r.h.s. of (1) would be odd and the denominator even, so $a$ could not be an integer. Hence $k$ is even.

If $k = 2$ then the cross-sum of $10^n + 2$ is 3, which is not divisible by $(2+1)^2 = 9$. The case $k = 4$ also leads to a contradiction, since $10^n + 4$ ends with 4, hence cannot be divisible by $(4+1)^2 = 25$. Thus $k \ge 6$.
In the following we show first that $k \le 8$ and finally that $k \ne 8$. The assumptions $k a = b \ge 10^{n-1}$ give $10 k a \ge 10^n$. Equality (1) implies
$$10^n = (k+1)^2 \cdot a - k = k^2 a + 2 k a + a - k = (k+2) \cdot k a + a - k.$$
Thus $10 k a \ge (k+2) \cdot k a + a - k$, whence
$$(8-k) \cdot k a \ge a-k.$$
As $a$ is positive, $(8-k) \cdot k a > -k$. As both sides of this inequality are divisible by $k$, this implies $(8-k) \cdot k a \ge 0$. Consequently $8-k \ge 0$, i.e., $k \le 8$.
If $k=8$, the inequality (2) implies $a \le 8$ whereas the equality (1) reduces to $a = \frac{10^n+8}{81}$. Hence $a$ ends with digit 8, leaving $a=8$ and $b=8 \cdot 8 = 64$ as the only possibility. But $864 \ne (8+64)^2$, contradicting the conditions of the problem.