As @StéphaneLaurent points out, $(X,Z)$ have a bivariate normal distribution and $E[X\mid Z] = aZ+b$. But even more can be said in this case because it is known
that
$$a = \frac{\operatorname{cov}(X,Z)}{\sigma_Z^2},
\quad b = \mu_X - a\mu_Z
= \mu_X - \frac{\operatorname{cov}(X,Z)}{\sigma_Z^2}\mu_Z,$$
and we can use the independence of $X$ and $Y$ (which
implies $\operatorname{cov}(X,Y) = 0$) to deduce that
$$\begin{align}
\operatorname{cov}(X,Z) &= \operatorname{cov}(X,X+Y)\\
&= \operatorname{cov}(X,X) + \operatorname{cov}(X,Y)\\
&= \sigma_X^2\\
\sigma_Z^2 &= \operatorname{var}(X+Y)\\
&= \operatorname{var}(X)+\operatorname{var}(Y) + 2\operatorname{cov}(X, Y)\\
&= \sigma_X^2+\sigma_Y^2\\
\mu_Z &= \mu_X+\mu_Y.
\end{align}$$
Note that the method used above can also be applied
in the more general case when $X$ and $Y$ are correlated
jointly normal random variables instead of independent normal random
variables.
Continuing with the calculations, we see that
$$E[X\mid Z] = \frac{\sigma_X^2}{\sigma_X^2+\sigma_Y^2}(Z-\mu_Z)
+ \mu_x \tag{1}$$
which I find comforting because we can interchange the roles
of $X$ and $Y$ to immediately write down
$$E[Y\mid Z] = \frac{\sigma_Y^2}{\sigma_X^2+\sigma_Y^2}(Z-\mu_Z)
+ \mu_Y\tag{2}$$
and the sum of $(1)$ and $(2)$ gives $E[X\mid Z] + E[Y\mid Z] = Z$
as noted in Stéphane Laurent's answer.
