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Let $ A, B $ be two square matrices of order $n$. Do $ AB $ and $ BA $ have same minimal and characteristic polynomials?

I have a proof only if $ A$ or $ B $ is invertible. Is it true for all cases?

If $A$ is invertible then $A^{-1}(AB)A= BA$, so $AB$ and $BA$ are similar, which implies (but is stronger than) $AB$ and $BA$ have the same minimal polynomial and the same characteristic polynomial.
The same goes if $B$ is invertible.

In general, from the above observation, it is not too difficult to show that $AB$, and $BA$ have the same characteristic polynomial, the type of proof could depends on the field considered for the coefficient of your matrices though.
If the matrices are in $mathcal{M}_n(mathbb C)$, you use the fact that $operatorname{GL}_n(mathbb C)$ is dense in $mathcal{M}_n(mathbb C)$ and the continuity of the function which maps a matrix to its characteristic polynomial. There are at least 5 other ways to proceed (especially for other field than $mathbb C$).

In general $AB$ and $BA$ do not have the same minimal polynomial. I'll let you search a bit for a counter example.

Before proving $AB$ and $BA$ have the same characteristic polynomials show that if $A_{mtimes n}$ and $B_{ntimes m} $ then characteristic polynomials of $AB$ and $BA$ satisfy following statement: $$x^n|xI_m-AB|=x^m|xI_n-BA|$$ therefore easily conclude if $m=n$ then $AB$ and $BA$ have the same characteristic polynomials.

Define $$C = begin{bmatrix} xI_m & A \B & I_n end{bmatrix}, D = begin{bmatrix} I_m & 0 \-B & xI_n end{bmatrix}.$$ We have
$$
begin{align*}
det CD &= x^n|xI_m-AB|,\
det DC &= x^m|xI_n-BA|.
end{align*}
$$
and we know $det CD=det DC$ if $m=n$ then $AB$ and $BA$ have the same characteristic polynomials.

Hint: Consider $A = begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix}$ and $B = begin{bmatrix} 0 & 0 \ 0 & 1 end{bmatrix}$. What do you get in that case?

It's not true that their characteristic polynomials will be the same in the general case. The best result in this general vein is the following.

Let $Ainmathbb{F}^{m times n}$ and let $Binmathbb{F}^{n times m}$, and $AB$, $BA$ with minimal polynomials (over $mathbb{F}$) $m_{AB}(x)$ and $m_{BA}(x)$ respectively. Then one of the following holds:

$m_{AB}(x) = m_{BA}(x)$, or $m_{AB}(x) = x cdot m_{BA}(x)$, or $xcdot m_{AB}(x) = m_{BA}(x)$.

It's easy, just use the fact that $(BA)^k=B(AB)^{k-1}A$.

For squre matrix, the charateristic polynomials are same, but for $A$ is matrix of size $m times n$ and B be matrix of size $n times m$. Then $x^{m}C_{BA}(x)=x^{n}C_{AB}(x)$ this implies that the nonzero eigenvalue of $AB$, counted with multiplicites, are same as nonzero eigenvalue of $BA$.

That is if A is of size 7×4 and B is of size 4×7 and assume that the 4×4 matrix BA has nonzero eigenvalue 1,1,3 so fourth eigenvalue of BA is 0. Then the 7×7 matrix AB will also have nonzero eigenvalue 1,1,3 and remaining four eigenvalue of AB are zero.