If you have more rows that columns, then all rows can NOT be linearly independent — but then, it actually does NOT matter here. You have the right answer in your second paragraph: if you know that all $N$ columns of the coefficient matrix are linearly independent, and if you know that solutions exist, then it has to be the case of a unique solution, since in this case
$$\operatorname{rank}(\text{coefficient matrix})=\operatorname{rank}(\text{augmented matrix})=\text{number of variables}.$$
The extra equations in this case don't affect the number of solutions. They just happen to be extraneous, or redundant, equations. They are linearly dependent on other equations, but that in and by itself does NOT create linear dependency among variables or solutions. Here's a quick example:
$$\left\{\begin{align} x&=5 \\ 2x&=10\end{align}\right.$$
You can see that in this case:
$$\operatorname{rank}(\text{coefficient matrix})=\operatorname{rank}(\text{augmented matrix})=\text{number of variables}=1,$$
so the system has a unique solution, even though $N=1$ and $L=2$. The second equation is just a redundant one, a multiple of the first equation, and it can be eliminated (it will become a row of zeroes when you do Gauss-Jordan reduction).
Also, regarding your statement:
… then the row rank of the matrix can be less than $N$.
The row and column ranks of a matrix are always equal. So if you already know that the column rank of a matrix is $N$, then its row rank can't possibly be less than $N$.
If you reduce the coefficient matrix, you should find that the last 2 equations are a multiple of the first, so the system of equations reduces to $$ x + 2y + 4z=0 $$
As you noticed, if $y=z=0$, then $x=0$. So $(0,0,0)$ is a particular solution of the system.
Now we ask the question, what if $y$ and $z$ were not $0$? We can answer that question by making $y\neq0$, then $z\neq0$. You will notice that the vectors we end up with are NOT multiples of each other, so they are "linearly independent"
1) If $y=c\neq0\in\mathbb{R}$ and $z=0$, then $x=-2c$ and $(-2c,c,0)=c(-2,1,0)$ is a solution for any $c\in\mathbb{R}$.
2)If $z=d\neq0\in\mathbb{R}$ and $y=0$, then $x=-4d$ and $(-4d,0,d)=d(-4,0,1)$ is a solution for any $d\in\mathbb{R}$.
As mentioned earlier, there does not exist a $c$ and $d$ such that $c(-2,1,0)+d(-4,0,1)=0$. Therefore the vectors are not multiples of each other. This condition is the definition of linear independence
Best Answer
Let $B^n_s$ be the set of of points in $n$ dimensions that have integer coords $\le s$ in modulus. Then $A$ takes $B^n_s$ into $B^m_t$ with $t=|A|ns$, with $|A|$ denoting the element of $A$ with the highest magnitude. Now $B^n_s$ has $(2s+1)^n$ elements, and $B^m_t$ has $(2|A|ns+1)^m$. If I make $s$ sufficiently large, the former exceeds the latter and so (pigeon-hole principle) two points must have the same image. Their difference maps to zero, and this gives a solution to $Ax=0$. This method also gives an upper bound on the size of such a solution. Proof courtesy of Lang.