Why do I see a frequency shift of the peak between the S11 and impedance response?
[This is more of an electronics related question and perhaps should be migrated to Electronics.SE?]
This can occur with a series parasitic inductance or capacitance in the resonator; the case with a series inductance is demonstrated below.
The Smith Chart visual provides immediate insight of a shunt L-C resonator (tank circuit) to ground. In the first plot where we have no parasitic inductance in series with an exaggerated parasitic resistance in the resonator (reducing the input impedance at resonance when the reactive inductance cancels the reactive capacitance). At resonance $X_{Lp}$ cancels $X_{Cp}$ and we would be at the large dot shown (with just the shunt resistance due to finite Q remaining, in parallel with the 50 ohm load and thus reducing the input impedance along the resistance axis). As we move off resonance lower in frequency the tank circuit is inductive and ultimately a short to ground at DC, following the shunt inductance trajectory as shown by the path marked $L_p$ indicating a parallel inductor to ground, as we sweep the test frequency toward DC terminating on the left origin in the Smith Chart consistent with a short circuit. As we go higher in frequency the tank circuit will be capacitive and ultimately a short to ground at $f=\infty$, following the shunt capacitor trajectory as shown by the path marked $C_p$ as we sweep the frequency higher.   In this case for all other frequencies other than at resonance the net impedance of the tank would smaller with a maximum tank impedance at resonance, and the resonance peak would match the S11 null.

Observe if the tank circuit had a series inductance effectively prior to the shunt L and C to ground. This inductor in series would follow the trajectory marked by $L_s$ (and would ultimately end at the right origin where $Z=\infty$, terminating at the large dot shown at resonance with $X_{Lp} = -X_{Cp}$. As we move off resonance upward in frequency, the resonator adds a shunt capacitance that in this case brings us closer to the center origin (where Z=50 ohms, perfect match), providing a lower S11 as it maximizes the net impedance of the resonator's connection to ground; essentially the resonator is tuning out the series inductance as it becomes capacitive. Not detailed here is that series inductance would also increase as frequency increases but we can see how if the capacitance relative to the inductance is significantly larger that the trajectory would move down faster than up and to the right resulting in the minimum as the OP is seeing. Further the inductance itself would have a point where it resonates out and becomes capacitive, so we could also easily envision the possible case of a net circular path of the series inductance and shunt capacitance of the resonator passing right through the center of the Smith chart (or closer to it).

Below is a basic simulation of this compliments of https://www.will-kelsey.com/smith_chart/
At resonance the Smith chart would be at DP5 (where the red line intersects the black line to the upper left of center. As we sweep frequency, the trajectory follows the red line where we see a minimum impedance is reached which is not on the resonance of the shunt L and C.