Let $n = y^2$ and $x$ be the number formed by the first 3 digits of $n$. Note that $y$ is a 3-digit number. Then
$$y^2 = 1000x + x + 1 = 1001x + 1 \quad \Rightarrow \quad (y-1)(y+1) = 7 \times 11 \times 13x.$$

Since $y \le 999$, not all of 7, 11, 13 can be factors of $y-1$. Thus we have 6 cases:

Case 1: $77 \mid y+1, 13 \mid y-1$. Then $y+1 = 77\alpha \le 1000$ and so $\alpha \le 12$. Testing $\alpha = 1, 2, \dots, 12$, we see that only $\alpha = 11$ works. We get $y = 846$ and $n = 715716$. The other 5 cases are $7 \times 13 \mid y+1, 11 \mid y-1$; $11 \times 13 \mid y+1, 7 \mid y-1$; $7 \mid y+1, 11 \times 13 \mid y-1$; $11 \mid y+1, 7 \times 13 \mid y-1$; $13 \mid y+1, 7 \times 11 \mid y-1$. They yield the solutions $n = 528529, 183184, 328329, 75076, 24025$. The 2 5-digit numbers are discarded and we have 4 solutions.