My original answer is below. I proceeded to validate the result with a quick simulation and immediately find some fundamental flaws with the premise. First I want to offer if the intention is for an impedance measurement comparing voltage to current, the "signal" waveforms should then be highly correlated (with phase rotation for complex impedance cases). For that purpose, and importantly assuming the noise itself is uncorrelated for the voltage and current measurements, I recommend a correlation based approach, which under the presence of additive white noise (and stable impedance over course of measurement) would provide the best estimate for the desired magnitude ratio as well as relative phase from which an accurate impedance can be derived (ideal for a Smith chart measurement if desired). This should be done on the time domain signals directly and not FFT magnitudes, which convert Gaussian distributions with zero mean to Rayleigh distributions, with the penalty of doing post-detection estimation. If there is further interest in the details of such a suggested approach, I suggest posting that as a separate question since that simply motivated the interesting stochastic processing question posed here.
Apart from that, I found the OP's challenge interesting with reference to MattL's original post showing the solution for multiplying two noisy signals, and proceeded with a generalized solution for dividing two noisy signals as the OP directed, with the conditions of independence and zero mean for the noise as well as the signals. I then proceeded with a validation of a simple case of Gaussian distributed signals as well as noise (to confirm the $K$ factor detailed below) and from that see the issue with the measurement as constructed:
If the signals were actually zero mean and independent, then the denominator consisting of signal and noise as a zero mean signal will pass through zero and approach zero, and for all small values the the quotient will explode! The measurement can't be done as described (but this will not pose a problem for my suggestion of a correlation approach). I should have seen this right away but it took my initial simulation to make that clear, with results shown in the plot below detailing the issue described above.
I am hoping the above provides a useful answer and guidance as to next steps toward a productive impedance measurement, and will leave the details below as an interesting exercise of using stochastics for further review and scrutiny. I don't see a mistake is the specific steps shown below other than now recognizing that the very first line will lead into an unmanageable signal given the divide by zero conditions), and for that reason am unable to verify the result.
Original response:
$$\frac{x_1 + n_1}{x_2+n_2} = \frac{x_1}{x_2}+\frac{n_1 x_2 - n_2 x_1}{x_2(n_2+x_2)} \tag{1}\label{1}$$
The signal component in the division result is
$$x = \frac{x_1}{x_2}\tag{2}\label{2}$$
And the noise is
$$ n =\frac{n_1 x_2 - n_2 x_1}{x_2(n_2+x_2)}\tag{3}\label{3}$$
Given independent zero mean signals, the signal power is:
$$\sigma_x^2 = \frac{\sigma_{x_1}^2}{\sigma_{x_2}^2}\tag{4}\label{4}$$
And the noise power is (note! the variance of the difference is the sum of the individual variances):
$$\sigma_n^2 = \frac{\sigma_{n_1}^2\sigma_{x_2}^2 +\sigma_{ n_2}^2\sigma_{ x_1}^2}{\sigma_{x_2}^2\sigma_{n_2}^2+K\sigma_{x_2}^4}\tag{5}\label{5}$$
Where $K\sigma_{x_2}^4$ represents the variance of $x_2^2$ and $\sigma_{x_2}^4 = (\sigma_{x_2}^2)^2$ (the variance of $x_2$, squared).
Note the factor of $K$: knowing the variance only for the signal $x_2$ is insufficient information to compute the variance of the product. For example, if we also knew the signal was Gaussian distributed then $K=2$ and if the signal was a sinusoid then $K=1.5$ (with "mean-square" instead of "variance"), but would be different for other distributions.
Proceeding with the SNR as ratio of signal to noise:
$$\text{SNR}= \frac{\sigma_x^2}{\sigma_n^2} = \frac{\sigma_{x_1}^2}{\sigma_{x_2}^2}\frac{\sigma_{x_2}^2\sigma_{n_2}^2+K\sigma_{x_2}^4}{\sigma_{n_1}^2\sigma_{x_2}^2 +\sigma_{ n_2}^2\sigma_{ x_1}^2}\tag{6}\label{6}$$
$$= \frac{ \sigma_{x_1}^2K\sigma_{x_2}^4+\sigma_{x_1}^2 \sigma_{x_2}^2\sigma_{n_2}^2}
{\sigma_{n_1}^2K\sigma_{x_2}^4+\sigma_{x_1}^2 \sigma_{x_2}^2\sigma_{n_2}^2 }\tag{7}\label{7}$$
Where:
$K\sigma_{x_2}^4$ represents the variance of $x_2^2$,
$K$ is a proportionality constant that depends on the distribution of signal $x_2$,
$\sigma_{x_2}^4 = (\sigma_{x_2}^2)^2$
Further simplifying \ref{7} as follows:
$$\text{SNR } = \frac{\sigma_{x_2}^2\sigma_{x_1}^2 ( K\sigma_{x_2}^2+\sigma_{n_2}^2)}
{\sigma_{x_2}^2(\sigma_{n_1}^2K\sigma_{x_2}^2+\sigma_{x_1}^2 \sigma_{n_2}^2 )} = \frac{ \sigma_{x_1}^2(K\sigma_{x_2}^2+\sigma_{n_2}^2)}
{\sigma_{n_1}^2K\sigma_{x_2}^2+\sigma_{x_1}^2 \sigma_{n_2}^2 }\tag{8}\label{8}$$
Which with $\text{SNR}_1 = \sigma_{x_1}^2/\sigma_{n_1}^2$ and $\text{SNR}_2 = \sigma_{x_2}^2/\sigma_{n_2}^2$ we can rewrite \ref{8} to be:
$$SNR = \frac{\text{SNR}_1(\text{SNR}_2K+1)}{\text{SNR}_2K+\text{SNR}_1}\tag{9}\label{9}$$
