If I'm just trying to show something is NP-hard (as opposed to NP-complete) does my reduction need to be completed in polynomial time?
Hardness/completeness is always defined with respect to a specific kind of reduction.

Using exponential-time reductions doens't work at all, because that means your reduction is even more powerful than the thing you're reducing to.*
Every language except $\emptyset$ and&nbsp;$\Sigma^*$ is NP-hard under exponential-time reductions. To see this, let $L$ be any language except $\emptyset$ and&nbsp;$\Sigma^*$, and let $X$ be in NP.  We can reduce $X$ to&nbsp;$L$ as follows. Fix two strings $w_{\mathrm{yes}}\in L$ and $w_{\mathrm{no}}\notin L$. Now, given a string $x$, we can determine in exponential time if $x\in X$. If it is, the reduction maps $x\mapsto w_{\mathrm{yes}}$; otherwise, it maps $x\mapsto w_{\mathrm{no}}$.

* OK, technically, we don't know that EXP is more powerful than NP, but it's certainly at least as powerful, and probably more powerful.