This equation has no solution. This is because, as a square number, $x^2$ can only end with digit $0$, $1$, $4$, $5$, $6$, and $9$. This means that the unit digit of $(x^2+3)$ can only be $3$, $4$, $7$, $8$, $9$, and $2$. it follows that $(x^2+3)$ is not a multiple of $5$. Now, we find the left side contains only one divisor of $5$. Hence, it cannot be a square number as the right side requires.