The "sampling theory" people will tell you that no such estimate exists. But you can get one, you just need to be reasonable about your prior information, and do a lot harder mathematical work.
If you specified a Bayesian method of estimation, and the posterior is the same as the prior, then you can say the data say nothing about the parameter. Because things may get "singular" on us, then we cannot use infinite parameter spaces. I am assuming that because you use Pearson correlation, you have a bivariate normal likelihood:
$$p(D|\mu_x,\mu_y,\sigma_x,\sigma_y,\rho)=\left(\sigma_x\sigma_y\sqrt{2\pi(1-\rho^2)}\right)^{-N}exp\left(-\frac{\sum_{i}Q_i}{2(1-\rho^2)}\right)$$
where
$$Q_i=\frac{(x_i-\mu_x)^2}{\sigma_x^2}+\frac{(y_i-\mu_y)^2}{\sigma_y^2}-2\rho\frac{(x_i-\mu_x)(y_i-\mu_y)}{\sigma_x\sigma_y}$$
Now to indicate that one data set may be the same value, write $y_i=y$, and then we get:
$$\sum_{i}Q_i=N\left[\frac{(y-\mu_y)^2}{\sigma_y^2}+\frac{s_x^2 + (\overline{x}-\mu_x)^2}{\sigma_x^2}-2\rho\frac{(\overline{x}-\mu_x)(y-\mu_y)}{\sigma_x\sigma_y}\right]$$
where
$$s_x^2=\frac{1}{N}\sum_{i}(x_i-\overline{x})^2$$
And so your likelihood depends on four numbers, $s_x^2,y,\overline{x},N$. So you want an estimate of $\rho$, so you need to multiply by a prior, and integrate out the nuisance parameters $\mu_x,\mu_y,\sigma_x,\sigma_y$. Now to prepare for integration, we "complete the square"
$$\frac{\sum_{i}Q_i}{1-\rho^2}=N\left[\frac{\left(\mu_y-\left[y-(\overline{x}-\mu_x)\frac{\rho\sigma_y}{\sigma_x}\right]\right)^2}{\sigma_y^2(1-\rho^{2})}+\frac{s_x^2}{\sigma_{x}^{2}(1-\rho^{2})} + \frac{(\overline{x}-\mu_x)^2}{\sigma_x^2}\right]$$
Now we should err on the side of caution and ensure a properly normalised probability. That way we can't get into trouble. One such option is to use a weakly informative prior, which just places restriction on the range of each. So we have $L_{\mu}<\mu_x,\mu_y<U_{\mu}$ for the means with flat prior and $L_{\sigma}<\sigma_x,\sigma_y<U_{\sigma}$ for the standard deviations with jeffreys prior. These limits are easy to set with a bit of "common sense" thinking about the problem. I will take an unspecified prior for $\rho$, and so we get (uniform should work ok, if not truncate the singularity at $\pm 1$):
$$p(\rho,\mu_x,\mu_y,\sigma_x,\sigma_y)=\frac{p(\rho)}{A\sigma_x\sigma_y}$$
Where $A=2(U_{\mu}-L_{\mu})^{2}[log(U_{\sigma})-log(L_{\sigma})]^{2}$. This gives a posterior of:
$$p(\rho|D)=\int p(\rho,\mu_x,\mu_y,\sigma_x,\sigma_y)p(D|\mu_x,\mu_y,\sigma_x,\sigma_y,\rho)d\mu_y d\mu_x d\sigma_x d\sigma_y$$
$$=\frac{p(\rho)}{A[2\pi(1-\rho^2)]^{\frac{N}{2}}}\int_{L_{\sigma}}^{U_{\sigma}}\int_{L_{\sigma}}^{U_{\sigma}}\left(\sigma_x\sigma_y\right)^{-N-1}exp\left(-\frac{N s_x^2}{2\sigma_{x}^{2}(1-\rho^{2})}\right) \times$$
$$\int_{L_{\mu}}^{U_{\mu}}exp\left(-\frac{N(\overline{x}-\mu_x)^2}{2\sigma_x^2}\right)\int_{L_{\mu}}^{U_{\mu}}exp\left(-\frac{N\left(\mu_y-\left[y-(\overline{x}-\mu_x)\frac{\rho\sigma_y}{\sigma_x}\right]\right)^2}{2\sigma_y^2(1-\rho^{2})}\right)d\mu_y d\mu_x d\sigma_x d\sigma_y$$
Now the first integration over $\mu_y$ can be done by making a change of variables $z=\sqrt{N}\frac{\mu_y-\left[y-(\overline{x}-\mu_x)\frac{\rho\sigma_y}{\sigma_x}\right]}{\sigma_y\sqrt{1-\rho^{2}}}\implies dz=\frac{\sqrt{N}}{\sigma_y\sqrt{1-\rho^{2}}}d\mu_y$ and the first integral over $\mu_y$ becomes:
$$\frac{\sigma_y\sqrt{2\pi(1-\rho^{2})}}{\sqrt{N}}\left[\Phi\left(
\frac{U_{\mu}-\left[y-(\overline{x}-\mu_x)\frac{\rho\sigma_y}{\sigma_x}\right]}{\frac{\sigma_y}{\sqrt{N}}\sqrt{1-\rho^{2}}}
\right)-\Phi\left(
\frac{L_{\mu}-\left[y-(\overline{x}-\mu_x)\frac{\rho\sigma_y}{\sigma_x}\right]}{\frac{\sigma_y}{\sqrt{N}}\sqrt{1-\rho^{2}}}
\right)\right]$$
And you can see from here, no analytic solutions are possible. However, it is also worthwhile to note that the value $\rho$ has not dropped out of the equations. This means that the data and prior information still have something to say about the true correlation. If the data said nothing about the correlation, then we would be simply left with $p(\rho)$ as the only function of $\rho$ in these equations.
It also shows how that passing to the limit of infinite bounds for $\mu_y$ "throws away" some of the information about $\rho$, which is contained in the complicated looking normal CDF function $\Phi(.)$. Now if you have a lot of data, then passing to the limit is fine, you don't loose much, but if you have very scarce information, such as in your case - it is important keep every scrap you have. It means ugly maths, but this example is not too hard to do numerically. So we can evaluate the integrated likelihood for $\rho$ at values of say $-0.99,-0.98,\dots,0.98,0.99$ fairly easily. Just replace the integrals by summations over a small enough intervals - so you have a triple summation
