Answer by periblepsis for Analyzing a DC circuit

almost completed labeling

Just for clarity (your image is far too fuzzy) and to graphically label the nodes you identified in your writing:

^{simulate this circuit– Schematic created using CircuitLab}

ground reference

All of these nodes cannot be left as undefined values. Take the following circuit, for example:

^{simulate this circuit}

Start at the bottom and go clockwise, assuming the current is non-zero. If so, then it must be the case that \$V_b\gt V_a\$ due to Ohm's Law and \$R_1\$. But it must also be the case due to Ohm's Law and \$R_2\$ that \$V_a\gt V_b\$. These are contradictory statements.

So we find that the only allowable possibility is that the current must be zero. It cannot be otherwise.

But even so, what's the value of \$V_a=V_b\$? All answers are allowed. So there's an infinite number of choices. The only way to remove that ambiguity is to assign \$0\:\text{V}\$ to either \$V_a\$ or \$V_b\$. As soon as that's done then the other has a single defined value.

We call the node assigned \$0\:\text{V}\$ by many names. One of them is ground or ground reference. But it is important to have one. You can assign it anywhere you want. But you do need to assign \$0\:\text{V}\$ to one and only one node.

There's a larger reason that applies to all non-trivial circuits, making exactly this same requirement for all such circuits, and it's called called the Rank-Nullity Theorem. But that's for another time.

completed labeling

So one of the nodes must be selected as \$0\:\text{V}\$:

^{simulate this circuit}

But KCL is never performed for known-voltage (or ground) nodes. Only if the node voltage is unknown is it kept and not removed from analysis.

So the circuit becomes:

^{simulate this circuit}

But we aren't done, yet.

A current source has \$\infty\$ impedance. They also have two ends. But in this case, \$\infty\$ impedance means you can split it into two parts, with one part pointing into a node (and coming out of nowhere) and one pointing away from a node (and going nowhere.)

^{simulate this circuit}

The gap between the two currents is just another representation of \$\infty\$ impedance in the original current source.

From the above, you know that \$V_4=V_5+I_1\cdot R_6\$ and that \$V_3=V_2-I_1\cdot R_4\$.

At this point, you should be able to see that there are only two unknown node voltages. Everything else has been defined as a specific value or a specific value relative to an unknown.

So the KCL equations are:

$$\begin{align*}\frac{V_2}{R_2}+\frac{V_2}{R_3}+I_1&=\frac{+1\:\text{V}}{R_2}+\frac{V_5}{R_3}\\\\\frac{V_5}{R_3}+\frac{V_5}{R_5}&=\frac{V_2}{R_3}+I_1\end{align*}$$

Note that \$R_4\$ and \$R_6\$ aren't present. That's because they don't affect the current leaving \$N_2\$ or entering \$N_5\$.

Once you solve that pair of equations, you can then work out \$V_3\$ and \$V_4\$ from \$V_2\$ and \$V_5\$.

question

I need to find the nodes, the branches, and the loops. I need theequations describing the circuit based on Kirchoff's laws.

On first blush, this sounds like you want to analyze this using both mesh and nodal, separately. But directed graph theory combines both in a single descriptive structure. So that's would be the answer if you are looking for an approach that combines both. (It can also readily show you the meshes.) You can look at my answer here to see if that's where you wanted to go. But I don't want to compose this problem in that fashion unless you say that's the goal. So I'll leave it here, for now.