We know that all resistors dissipate at a rate $I^2 R = V^2 /R$.
But I'm wondering if it is possible to prove that a component with a linear relationship between current and voltage ($ I \propto V$) must dissipate at a rate $\propto I^2$ on entirely thermodynamic arguments.
Answer
The Onsager relations in thermodynamics state that the rate of entropy production (which can be related to the power dissipated at constant temperature) is proportional to the product of the flow (here, the current flux $I/A$ for a cross-sectional area of $A$) and the field (here, the electric field $V/L$ for a conductor length of $L$).
Since you've stated that Ohm's Law holds (i.e., $V\propto I$), then the electric field $V/L$ must scale with the current flux $I/A$ (and the resistivity). Assuming that the dimensions stay constant, the heat dissipated is thus proportional to $I^2$.
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