The last digit of a perfect square can be $0$, $1$, $4$, $5$, $6$ or $9$. Denote the given number by $n$. If $n$ is to be a perfect square it has to end in $0$.

Assume that the last $k$ digits of $n$ are equal to $0$ and the digit just to the left of them is not $0$. If $k$ is odd, $k = 2m - 1$. Then $\frac{n}{10^{2m-2}}$ is also a perfect square. It is divisible by $10$ so it must also be divisible by $25$. This is not possible since the second digit from the right is either $3$ or $7$.

Hence, $n$ must end in an even number of zeros and $k$ is even. In this case $\frac{n}{10^k}$ is a positive integer and a perfect square with the last digit equal to either $3$ or $7$. Again, this is not possible. Hence, $n$ is never a perfect square.