This is a pretty problem. I asked a question about it in MathOverflow a while back. Here is the solution using complex analysis, copying from the link above, and adding some details: (I've added the details in between square brackets. Ignore those paragraphs if you want to work out the details on your own.)
Assign to each progression $A_i=(a_i+kb_i\mid k\in{\mathbb N})$, $1\le i\le n$, its generating series, $f_i(x)=\sum_{k=0}^\infty x^{a_i+kb_i}$. Then $f_i(x)=x^{a_i}/(1-x^{b_i})$. Note the series converges for $|x|<1$.
[ To see this: $\sum_{k=0}^\infty x^{a_i+kb_i}=x^{a_i}\sum_{k=0}^\infty (x^{b_i})^k$, and use that if $|t|<1$, the geometric series $\sum_{k=0}^\infty t^k$ equals $1/(1-t)$. ]
Now, since the $A_i$ partition ${\mathbb N}$, we have $\sum_{i=1}^n f_i(x)=\sum_{m=1}^\infty x^m=x/(1-x)$.
[ For this, observe that if $m\in A_i$, then $x^m$ is one of the terms being added in $f_i(x)$, and $x^m$ is not a term in any of the other series $f_j$ for $j\ne i$. ]
If all the $b_i$ are different, let $b$ be the largest, and fix a primitive $b$-th root of unity $\zeta$. Now let $x\to\zeta$ to reach a contradiction.
[ In detail, $\zeta=e^{2\pi i/b}$ is a primitive $b$-th root of unity ($i$ the square root of $-1$). This means that $\zeta^b=1$, if $k$ is an integer, then $\zeta^k=1$ iff $k$ is a multiple of $b$, and if $z^b=1$, then $z=\zeta^n$ for some integer $n$. You then have that $$\frac{x^{a_1}}{1-x^{b_1}}+\dots+\frac{x^{a_n}}{1-x^{b_n}}=f_1(x)+f_2(x)+\dots f_n(x)=\sum x^m=\frac x{1-x}. $$
Call $(*)$ this equation. When $x\to\zeta$, we obtain on the right hand side of $(*)$ the fraction $\displaystyle \frac{\zeta}{1-\zeta}$. Note here that $\zeta\ne 1$ because we are assuming we have at least two arithmetic progressions, so $b>1$. On the other hand, if $b_j\ne b$, then $$ \frac{x^{a_j}}{1-x^{b_j}}\to\frac{\zeta^{a_j}}{1-\zeta^{b_j}} $$ and $\zeta^{b_j}\ne 1$ because $b>b_j$. But if $b_j=b$, then $\displaystyle \frac{x^{a_j}}{1-x^{b_j}}\to\infty$ because $\zeta^{b_j}=\zeta^b=1$. The contradiction is that the left hand side of $(*)$ converges to $\infty$, while the right hand side does not. ]
This shows that the largest of the common differences must appear at least twice.
As Gerry Myerson pointed out in MO, this argument is due to D J Newman. It appears in his book A Problem Seminar, problem 90, on page 18, with solution on page 100.
Here is an argument in the $m = 3$-case. What is interesting about it is that it shows that $n_{u, 3}$ is divisible by $n_{u, 1}$ at which point the $m = 3$-case follows from your treatment of the $m = 1$-case. It would be great if for all $m \geq 3$ we could find an $m' < m$ such that $n_{u, m'}$ divides $n_{u, m}$ but at present I don't know if that is the case.
So the $m=3$ argument. This is inspired by a now deleted post by someone who treated the $0_{u, 3}$ case.
Let $T_k$ denote the $k$'th triangular number. It is well known that the sum of the first $k$ third powers equals $T_k^2$. It follows that $n_{u, 3} = T_{n+u}^2 - T_{n-1}^2 = (T_{n+u} - T_{n-1})(T_{n+u} + T_{n-1})$.
Look at the first term in this factorization, $T_{n+u} - T_{n-1}$. On the one hand it is a divisor of the full thing, so of $n_{u, 3}$. Thus, if the latter is a power of two so is the former. On the other hand, $T_{n+u} - T_{n-1}$ equals $n_{u, 1}$.
Conclusion: if $n_{u, 3}$ is a power of 2, so is $n_{u, 1}$ which you already showed impossible.
Best Answer
There is no $(n,d,u,t)$ such that $$\sum_{q=0}^{u}(n+qd)^{2}=3^t\tag1$$
Proof :
Suppose that there is $(n,d,u,t)$ satisfying $(1)$ which is equivalent to $$(u+1)\left(6n^2+6ndu+u(2u+1)d^2\right)=2\cdot 3^{t+1}$$
Let us separate it into three cases. Note here that $u+1\gt 1$ and $6n^2+6ndu+u(2u+1)d^2\gt 2$.
Case 1 : $(u+1,6n^2+6ndu+u(2u+1)d^2)=(2,3^{t+1})$. Then,$$n^2+(n+d)^2=3^t\tag2$$We see that both $n$ and $n+d$ have to be divisible by $3$, so setting $n=3n_1$ and $n+d=3d_1$ gives$$n_1^2+d_1^2=3^{t-2}\tag3$$Comparing $(3)$ with $(2)$, we see that if $t$ is even, then there are positive integers $N,D$ such that $N^2+D^2=1$ which is impossible. If $t$ is odd, then there are positive integers $N,D$ such that $N^2+D^2=3$ which is impossible.
Case 2 : $(u+1,6n^2+6ndu+u(2u+1)d^2)=(2\cdot 3^a,3^b)$ where $a,b$ are positive integers such that $a+b=t+1$. Then,$$6n^2+6nd(2\cdot 3^a-1)+(2\cdot 3^a-1)(4\cdot 3^a-1)d^2=3^b\tag4$$Setting $d=3d_1$ gives$$2n^2+2n\cdot 3d_1(2\cdot 3^a-1)+(2\cdot 3^a-1)(4\cdot 3^a-1)\cdot 3d_1^2=3^{b-1}$$Setting $n=3n_1$ gives$$6n_1^2+6n_1d_1(2\cdot 3^a-1)+(2\cdot 3^a-1)(4\cdot 3^a-1)d_1^2=3^{b-2}\tag5$$Comparing $(5)$ with $(4)$, we see that if $b$ is odd, then there are positive integers $a,N,D$ such that$$6N^2+6ND(2\cdot 3^a-1)+(2\cdot 3^a-1)(4\cdot 3^a-1)D^2=3$$which is impossible since the LHS is larger than $3$. If $b$ is even, then there are positive integers $a,N,D$ such that $$6N^2+6ND(2\cdot 3^a-1)+(2\cdot 3^a-1)(4\cdot 3^a-1)D^2=1$$which is impossible since the LHS is larger than $1$.
Case 3 : $(u+1,6n^2+6ndu+u(2u+1)d^2)=(3^a,2\cdot 3^b)$ where $a,b$ are positive integers such that $a+b=t+1$. Then, $$6n^2+6nd(3^a-1)+(3^a-1)(2\cdot 3^a-1)d^2=2\cdot 3^b\tag6$$Setting $d=3d_1$ gives$$2n^2+2n\cdot 3d_1(3^a-1)+(3^a-1)(2\cdot 3^a-1)\cdot 3d_1^2=2\cdot 3^{b-1}$$Setting $n=3n_1$ gives$$6n_1^2+6n_1d_1(3^a-1)+(3^a-1)(2\cdot 3^a-1)d_1^2=2\cdot 3^{b-2}\tag7$$Comparing $(7)$ with $(6)$, we see that if $b$ is odd, then there are positive integers $a,N,D$ such that$$6N^2+6ND(3^a-1)+(3^a-1)(2\cdot 3^a-1)D^2=6$$which is impossible since the LHS is larger than $6$. If $b$ is even, then there are positive integers $a,N,D$ such that $$6N^2+6ND(3^a-1)+(3^a-1)(2\cdot 3^a-1)D^2=2$$which is impossible since the LHS is larger than $2$.
From the three cases above, the conclusion written at the top follows.