Let this $4$-digit number be $N$, then $N = k^2$ for some integer $k$. Increasing every digital of $N$ by $1$ is the same as adding $1111$ to $N$. Hence, $(N+1111)$ is also a square. Let it be $m^2$. Subtracting these two relations leads to $$(N+1111)-N = m^2 - k^2 \implies 1111 = (m+k)(m-k)$$

There are only two ways to express $1111$ as a multiple of two integers: $1\times 1111$ and $11\times 101$. Therefore we have $$m+k = 1111, m-k = 1$$ or $$m+k = 101, m-k=11$$

Solving these two systems and noting $k^2$ is a $4$ digit-number leads to only one solution $k=45$ which means $N=\boxed{2025}$.