Others have hit on this, but I thought I'd contribute how I'd write the problem down (briefly):
Let $x_i \sim N(m, 1)$ for $i \in \lbrace 1, \dots, N\rbrace$ and define $y_i \equiv x_i$ if $x_i > 0$ and $y_i \equiv 0$ if $x_i \leq 0$. The data in this case is the sample $Y = \lbrace y_1, \dots, y_N \rbrace$. So any likelihood based method (Bayes, MLE, etc) starts with the likelihood for $Y$. m$ (or sampling model for $Y$ depending on how you want to look at it). Letting $n$ denote the number of exactly zero observations, this likelihood can be simply expressed as $$\Phi(-m)^n \times \prod_{j \;\mid \; y_j \;\neq \;0} N^+(y_j \;;\; m, 1),$$ where $\Phi(\cdot)$ is the standard Normal CDF and $N^+(\ \cdot \mid \;mean, variance)$ denotes a positive truncated Normal pdf. By explicitly writing the normalizing constant, we can rewrite this as $$\frac{\Phi(-m)^n }{\left(1-\Phi(-m)\right)^{N-n}} \times \prod_{j \;\mid \; y_j \;\neq \;0} N(y_j \;;\; m, 1).$$ This expression makes clear why the zeros $do$ matter, because they are informative about $m$ via the ratio of zeros to non-zeros observed.
