You are in the right direction. From the formulation you have, observe that we can rewrite it as
\begin{align}v_1 &= \Delta_{11}u_1 \\ v_2 &= \Delta_{12}u_1+\Delta_{22}u_2 \\ v_2 &= \Delta_{13}u_1+\Delta_{23}u_2 + \Delta_{33}u_3 \\ &\dots~\dots~\dots\end{align} where I define $$\Delta_{ij} = {<v_i,u_j> \over <u_j,u_j>}~,~\forall i\leq j~,~\{\mbox{0 elsewhere}\}$$
Now try to see if $\Delta$ is the upper triangular matrix you are looking for.
I think our OP H_1317's proof is conceptually correct.
Here is a more somewhat more abstract proof:
Suppose $A$ is upper triangular; then I claim that $A^{-1}$ is also upper triangular; for we may write
$A = D + T, \tag 1$
where $D$ is diagonal and $T$ is strictly upper triangular; that is, the diagonal entries of $T$ are all zero; we observe that, since $A$ is triangular, $\det(A)$ is the product of the diagonal entries of $A$; since $A$ is unitary, it is non-singular and thus $\det(A) \ne 0$, so none of the diagonal entries of $A$ vanish, and the same applies to $D$; therefore $D$ is invertible and we may write
$A = D(I + D^{-1}T); \tag 2$
we next observe that $D^{-1}T$ is itself strictly upper triangular, hence nilpotent; in fact we have
$(D^{-1}T)^n = 0, \tag 3$
where $n = \text{size}(A)$; the nilpotence of $D^{-1}T$ allows us to write an explicit inverse for $I + D^{-1}T$; indeed, we have the well-known formula
$(I + D^{-1}T) \displaystyle \sum_0^{n - 1} (-D^{-1}T)^k = I + (-1)^n (D^{-1}T)^n = I; \tag 4$
thus,
$(I + D^{-1}T)^{-1} = \displaystyle \sum_0^{n - 1} (-D^{-1}T)^k; \tag 5$
since every matrix $(-D^{-1}T)^k$ occurring in this sum is upper triangular, we see that $(I + D^{-1}T)^{-1}$ is upper triangular as well; from (2),
$A^{-1} = (I + D^{-1}T)^{-1}D^{-1}, \tag 6$
which shows that $A^{-1}$ is upper triangular.
Having established my claim, we now invoke the unitarity of $A$:
$A^\dagger A = AA^\dagger = I, \tag 7$
i.e.,
$A^\dagger = A^{-1}; \tag 8$
we have by definition
$A^\dagger = (A^\ast)^T = ((D + T)^\ast)^T = (D^\ast + T^\ast)^T = (D^\ast)^T + (T^\ast)^T, \tag 9$
from which we see that $A^\dagger$ is lower triangular when $A$, and hence $A^{-1}$, is upper triangular;
then only way (8) can hold is with $T = 0$; therefore we see that
$A = D \tag{10}$
is a diagonal matrix.
Of course, if $A$ is lower triangular the same result binds, the proof almost identical to that given above. $OE\Delta$.
Best Answer
Your approach is OK. As you said, the columns of $A$, denote them by $C^A_1,...,C^A_n$, form an orthonormal basis of $F^n$ w.r.t the standard inner product. Denote $C^A_j=(a_{1,j},...,a_{n,j})$.
Since $A$ is upper-triangular, $a_{k,1}=0$ for all $k>1$. Since $A$ is unitary, $a_{1,1}\neq0$. Now compute the inner-product $\langle C^A_1,C^A_j\rangle$ for all $j>1$: you get: $$0=\langle C^A_1,C^A_j\rangle=a_{1,1}\cdot a_{1,j}+0+...+0$$ Since $a_{1,1}\neq0$, we have $a_{1,j}=0$ for all $j>1$. (this shows that the first row is $(a_{1,1}\hspace{2pt}0\hspace{2pt}...\hspace{2pt}0)$)
Now use induction on columns.