Does a^m * b^n = (ab)^m+n

no it cant be simplified

I think you're missing a pair of parentheses; didn't you mean

a^m * b^n =? (ab)^(m+n) ?

Because, as you've written it, the rules of precedence make the RHS =
[(ab)^m] + n
and that is clearly not = LHS.

With that in place, the answer is, only in a few special cases; in general, certainly not.
Take for instance, the case a=b:
LHS = a^m * b^n = a^m * a^n = a^(m+n)
RHS = (a²)^(m+n) = a^(2m+2n) = LHS²

Or take the case m=n:
LHS = a^m * b^n = a^n * b^n = (ab)^n
RHS = (ab)^(m+n) = [LHS] * (ab)^m

So one special case where it works is m=n=0.
Another is m=n and ab=1.
Another is a=b and m+n=0.

Another is a=b=1:
1^m * 1^n = 1 = 1^(m+n)

Another is a=0 or b=0:
0^m * b^n = 0 = 0^(m+n)

Another is a=1 and m=0:
1^0 * b^n = b^n = (1*b)^(0+n)

Likewise for b=1 and n=0.

But in the vast majority of cases, it won't hold.
Just try a=b=m=n=2; or a=2, b=m=n=1.

no.. it doesnt follow the rule
you should have same base in order to add their exponents b^m * b^n = b^m+n

let's test it
3^1*4^2=(3*4)^(1+2)
3*16=12^3
48=12*12*12
no