I know and I have already did the proof by considering an $n\times n$ arbitrary matrix and checking the conditions and figured it out that the matrix should be the identity matrix. Moreover, when I was searching here, I saw this question and its answers. The answers there are as well by doing calculations on matrices.
I would like if there is some more professional approach to do this proof.
First, it's easy to see that an upper-triangular unitary matrix must be diagonal:
If $A$ is upper-triangular the eigenvalues are exactly the entries on the diagonal. If $A$ is unitary the eigenvalues have absolute value $1$. And the rows are orthonormal, in particular each row has euclidean norm $1$. Since there's an entry of modulus $1$ on the diagonal all the other entries in that row must be $0$. So $A$ is diagonal.
And now if the diagonal entries are positive they must equal $1$, since that's the only positive number of modulus $1$. So $A=I$.
