I already found that \(f_{X,Z}(x,z)=\frac{1}{2\pi\sigma}e^{-\frac{1}{2\sigma^2}[(x-\mu)^2+\sigma^2(x-z)^2]}\) is bivariate normal and \(Z \sim N(\mu,\sigma^2+1)\)

\(f_{X|Z}(x|z)=\frac{f_{X,Z}(x,z)}{f_Z(z)}=\frac{\frac{1}{2\pi\sigma}e^{-\frac{1}{2\sigma^2}[(x-\mu)^2+\sigma^2(x-z)^2]}}{\frac{1}{\sqrt{2\pi}\sqrt{\sigma^2+1}}e^{-\frac{1}{2(\sigma^2+1)}(z-u)^2}}\)

\(f_{X|Z}(x|z)=\frac{\sqrt{\sigma^2+1}}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2\sigma^2}[(x-\mu)^2+\sigma^2(x-z)^2]+\frac{1}{(2\sigma^2+1)}(z-\mu)^2}\)

After some simplifications I get

\(f_{X|Z}(x|z)=\frac{\sqrt{\sigma^2+1}}{\sqrt{2\pi}\sigma}e^{\frac{1}{2\sigma^2(\sigma^2+1)}[-(\sigma^2+1)(x^2-2\mu x+\sigma^2(x^2-2zx)-2\sigma^2\mu z]}\)

from here I'm stuck.Someone said that the conditional distribution would be \(N(Z,1)\) but not give details.