You misunderstood the "axiom"
An Ideal opamp will do what it can to make the differential voltage between the -ve pin and the +ve pin equal zero. It doesn't state it is zero
All it can do is alter its output and thus with negative feedback there OPAMP stands a chance of making the difference zero
You then make use of that fact during circuit analysis to simplify calculations "if the difference is zero and the +ve terminal is at 0V, then the output must be...)
Thevenin's theorem is extremely important in practice. Every time an engineer talks about input or output impedance (or resistance, if you are concerned only in DC) it is applying that theorem implicitly.
Moreover, the theorem is applied whenever an engineer tries to estimate the input or output impedance of a complex circuit, where many components are connected together. Usually the application of the theorem is done visually, without doing explicit calculations if the circuit is relatively simple. But in some cases some back-of-the-envelope calculations may be needed.
Think of this common example: any time you speak about the internal resistance of a battery cell you are implicitly applying Thevenin's theorem. There is no actual resistor in a cell, besides the tiny resistance of the connection wires. The internal resistance is Thevenin's way to model the resistance of the connection wires together with all the resistive phenomena that take place in the chemistry of the battery.
For the same reasons its dual twin, Norton's theorem, is used (a bit less commonly).
And just to tell you something that is seldom written in basic circuit analysis books, although those theorems are strictly valid only if the circuit being substituted is linear, they are indeed applied also to non-linear circuits, if appropriate conditions are met.
Take as an example the output of a logic gate. You will see that sometimes, when discussing fan-out issues or when analog stuff is connected to that output, we talk about the gate output impedance, which in itself should be taboo since gates are strongly non-linear devices! But if you consider the two separate situations of when the output is high and when the output is low, you can talk about the output impedance, which of course could be different in the two situations!
What an engineer seldom uses in practice, especially nowadays in the era of SPICE simulators, are all those methods for calculating all the electrical quantities in a circuit, like nodal or mesh analysis and the like. Although an engineer should know about them for theoretical reasons, spending too much time in becoming the master of nodal analysis resolution is not time well spent. Much better to learn how to apply Thevenin's and Norton theorems (and a couple of other tricks) at a glance even in complicate circuits.
If you really want to understand how a real design engineer thinks, try the excellent book of Horowitz and Hill: The Art of Electronics. A bit on the expensive side, but it's a 1200 page monster covering most aspects of electronic design and it is written in a very easy-to-read style.
Best Answer
Let there be some linear circuit in a black box with two terminals exposed.
We measure the open circuit voltage \$V_{OC}\$ and the short circuit current \$I_{SC}\$ at the two terminals.
Now, according to Thevenin's theorem, the voltage across the terminals is given by
$$V_O = V_{OC} - I_O \frac{V_{OC}}{I_{SC}} = V_{TH} - I_OR_{TH}$$
Remember, this result is derived without an assumption as to the functional IV relationship for the load.
The above equation for \$V_O\$ is a load line and, whether the load is linear or not, the solution is the intersection of the load line and the IV curve of the load.
For example:
In equation form, let the load be some not necessarily linear circuit element where
$$V_L = V_O = f(I_L) $$
Then, it follows that
$$f(I_L) = V_{TH} - I_L R_{TH}$$