Prove that $\mathbb Z^n$ is not isomorphic to $\mathbb Z^m$ for $m\neq n$ 
Prove that $\mathbb Z^n$ is not isomorphic to $\mathbb Z^m$ for $m\neq n$.

My try:
Let $\mathbb Z^n\cong \mathbb Z^m $. To show that $m=n$.
Case 1: Let $m>n$. Now that $\mathbb Z^m$ has $m$ generators whereas $\mathbb Z^n$ has $n$  generators and an isomorphism takes a generator to generator ; that is the contradiction.
Please correct me if I am wrong.
The case  $m<n$ also follows similarly.

First proof
Suppose $\mathbb{Z}^n\cong\mathbb{Z}^m$, with $m\ne n$. Then you can find a set of $m$ free generators in $\mathbb{Z}^n$, say $\{f_1,f_2,\dots,f_m\}$. Then, for some integers $a_{ij}$, $b_{ij}$, you have
\begin{align}
f_j&=\sum_{i=1}^n a_{ij} e_i &&j=1,\dots, m\\
e_i&=\sum_{j=1}^m b_{ij} f_j &&i=1,\dots, n
\end{align}
where $\{e_1,\dots,e_n\}$ is the standard basis.
It's easy to show from this that the matrix $[a_{ij}]$ is invertible (over the integers, so, a fortiori, over the reals). But no matrix with a different number of rows and columns can have both a left and a right inverse.
Second proof
If $\mathbb{Z}^n\cong\mathbb{Z}^m$, then
$$
\mathbb{Z}^n\otimes_{\mathbb{Z}}\mathbb{Q}\cong
\mathbb{Z}^m\otimes_{\mathbb{Z}}\mathbb{Q}
$$
so
$$
\mathbb{Q}^n\cong\mathbb{Q}^m
$$
which implies $n=m$ by well known facts of linear algebra.
Proof of the more general result for any commutative ring
Let $A$ be a nontrivial commutative ring (with identity) and suppose $A^n\cong A^m$ as $A$-modules. Let $I$ be a maximal ideal of $A$; then
$$
A^n\otimes_A(A/I)\cong A^m\otimes_A(A/I)
$$
that immediately gives
$$
(A/I)^n\cong(A/I)^m
$$
as $(A/I)$-modules. Since $A/I$ is a field, we get $n=m$.
If, instead of a maximal ideal we take a prime ideal $P$ and tensor with the field of quotients of $A/P$, we get our second proof for the special case of a domain (with $(0)$ as the prime ideal).
