Is supermesh the right way to go?
You have been here long enough, I think, to be aware of the schematic editor. It's helpful (saves me time, anyway) if you bother, because the parts get labeled and it's easier to ask you clarifying questions. (But if all you have is a cell phone, it may be the case that you can only add pictures.) That said, here's the schematic I wish you'd drawn:







simulate this circuit &ndash; Schematic created using CircuitLab

I've kept the general format you provided, but as you can see parts are labeled and so are a few nodes. You may notice that I've selected one of the nodes as "ground." You get to do this for exactly one node of your choice. I picked that one, but you could choose a different one if you want. It won't change the answers to the questions. Still, I like my choice better than others. So that's my call.

One thing you should remember is that the impedance of a current source is infinite. So \$I_1\$ in series with \$R_2\$ as shown above means that you can just short out \$R_2\$ (remove it, replacing it with a wire.) That's because \$R_2\$'s value is vanishingly small compared to the theoretical impedance of \$I_1\$. So it contributes nothing to the analysis.

Another way to see this is that although \$I_1\$ causes a \$5\:\text{V}\$ drop across \$R_2\$, you don't know the voltage across \$I_1\$ which can be anything. So the voltage across \$I_1\$ just "adapts." Inserting a resistor just means it adapts to a different voltage value. But the summed voltage drop across the series pair, \$I_1\$ and \$R_2\$, will be the same no matter what the value of \$R_2\$ is.

We can rewrite the above schematic, now, like this:







simulate this circuit

Now, this isn't all you can simplify. \$I_1\$, no matter its value, has zero impact on the node marked as \$+4\:\text{V}\$. There is a voltage source that is holding that node's value. So we can re-write the schematic as follows, without impacting any of the analysis you require:







simulate this circuit

By now, if not before, I think you can see that the two current sources connected to node A result in a net current that must flow through \$R_3\$ (due to KCL.) So you should have a very firm grip on the value of \$I_3\$ at this point. It just "falls out," if it hadn't beforehand.

You can how replace \$R_3\$ with that value of current (which I'm sure you can easily find.)







simulate this circuit

At this point, all you need to do is work out the voltage value of node B and you will be able to work out the other two currents.

This is a very simple equation using KCL:

$$\frac{V_\text{B}}{R_1}+\frac{V_\text{B}}{R_4}=I_3+\frac{V_1=+4\:\text{V}}{R_1}$$

I think this should allow you to completely resolve the circuit without any worries over simultaneous equations. Nothing more than a bit of simple algebraic manipulation of a single equation and you'll have all you need to solve for the remaining two currents.