How do I convert benzoic acid to m-nitrobenzyl alcohol using 2 steps?
Use the meta-directing effect of $\ce{-COOH}$: nitrate benzoic acid first. 
Reduce m-nitrobenzoic acid with $\ce{LiAlH4}$. The nitro group is not attacked, but when you calculate the equivalents of  $\ce{LiAlH4}$ needed, don't forget that you are reacting an acid with a hydride.



Edit 1:
Nitroalkanes or nitroalkyl-substituted arenes can be converted to the corresponding amines by $\ce{LiAlH4}$. This is usually not the case for nitroarenes. Here, catalytic hydrogenation, e.g. over $\ce{Pd/C}$ is the method of choice.

Edit 2:
Let's play, change the order of events and first reduce benzoic acid to benzyl alcohol. The hydroxymethyl group isn't m-directing, but what will happen to it under nitration conditions ($\ce{H2SO4}$ and $\ce{HNO3}$)?

Under these pretty acidic conditions, elimination of water from the starting material furnishes benzyl cations, which might react with benzyl alcohol in different ways, such as (multiple) Friedel-Crafts alkylations or etherification. I'd dare to say that the result is a mess.