Already if an integer is a square it must be of the shapes you have described.
It is easiest to prove this using congruence notation. Any integer is congruent to $0$, $1$, $2$, $3$, or $4$ modulo $5$. Note now that $0^2$, $1^2$, $2^2$, $3^2$, and $4^2$ are respectively congruent to $0$, $1$, $4$, $4$, and $1$ modulo $5$, so never to $2$ or to $3$ modulo $5$.
If we do not want to use congruence notation, calculate the remainder when $(5k)^2$, $(5k+1)^2$, $(5k+2)^2$, $(5k+3)^2$, and $(5k+4)^2$ are divided by $5$. As a sample, let's do it for $(5k+3)^2$. This is $25k^2+30k+9$, which is $5(5k^2+6k+1)+4$, so it has remainder $4$ on division by $5$.
Out of curiosity, let us explore the cubes modulo $5$. Note that $0^3$, $1^3$, $2^3$, $3^3$, and $4^3$ are congruent respectively to $0$, $1$, $3$, $2$, and $4$ modulo $5$. So, unlike squares, cubes can take on any value modulo $5$.
By looking at the numbers $0$, $1$, and $64$, which are all perfect squares and perfect cubes, we can see that indeed a number which is a perfect square and a perfect cube can be of any of the shapes described.