That is a false assumption since a (nXn matrix) a square matrix needs to have at least n different eigenvalues (to make eigenvectors from)
No, they need not be different, as you yourself have noticed (the identity matrix example) and others have also pointed it out.
You can write each matrix in the Jordan normal form:
$$A = S J S^{-1}, \quad J = \operatorname{diag}(J_1, J_2, \dots, J_m),$$
where
$$J_k = \begin{bmatrix} \lambda_k & 1 \\ & \ddots & \ddots \\ & & \lambda_k & 1 \\ & & & \lambda_k \end{bmatrix}.$$
A matrix $A$ is diagonalizable if and only if all $J_k$ are of order $1$, i.e., $J_k = \begin{bmatrix} \lambda_k \end{bmatrix}$. In other words, $m = n$ and
$$J = \operatorname{diag}(\lambda_1, \lambda_2, \dots, \lambda_n).$$
Now, if $A$ has only one eigenvalue, that means that $\lambda := \lambda_1 = \lambda_2 = \cdots = \lambda_n$, so
$$J = \operatorname{diag}(\lambda, \lambda, \dots, \lambda) = \lambda \operatorname{diag}(1, 1, \dots, 1) = \lambda {\rm I}.$$
Now, let us get back to $A$:
$$A = S J S^{-1} = S (\lambda {\rm I}) S^{-1} = \lambda S S^{-1} = \lambda {\rm I}.$$
So, $A$ is diagonalizable with only one eigenvalue if and only if it is a scalar matrix.
However, not every matrix with only one eigenvalue is diagonalizable. Just put $\lambda := \lambda_1 = \lambda_2 = \cdots = \lambda_m$, but let some of $J_k$ be of order strictly bigger than $1$. For example,
$$B = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$
is not diagonalizable.
The action of $B$ on $v_j$ is
$$ Bv_j = (A - \lambda_i v_iv_i^T)v_j = Av_j - \lambda_i v_i v_i^T v_j $$
Since the eigenvectors are orthogonal, then $v_i^Tv_j = \delta_{ij}$, assuming the eigenvectors are suitably normalized. For $j\ne i$, then $Bv_j = A v_j = \lambda_j v_j$. Therefore $\lambda_j$ is an eigenvalue of $B$ when $j\ne i$, and the corresponding eigenvector is the same $v_j$.
When $j=i$, then
$$ Bv_i = Av_i - \lambda_i v_i = \lambda_i v_i - \lambda_i v_i = 0 $$
so $\lambda_i$ of $B$ is zero.
Best Answer
Hint for one direction: If $A$ is diagonalizable as $A=PDP^{-1}$, then the characteristic polynomial of $A$ is the same as the characteristic polynomial of $D$ (why?). The latter is easy to compute and allows you to describe the algebraic multiplicities. The geometric multiplicities are also easy to describe, since you have all the eigenvectors (columns of $P$).
Hint for the other direction: if all the geometric and algebraic multiplicities match up, then you are able to construct an eigenbasis for $\mathbb{R}^n$ and directly build your diagonalization.