There is no finite ring with $34$ units (I assume all rings are commutative, with $1$). First, recall the following fact about local rings:
$\newcommand{\char}{\operatorname{char}}$
$\newcommand{\m}{\mathfrak{m}}$
$\newcommand{\Z}{\mathbb{Z}}$

**Proposition:** Let $(R,\m)$ be a local ring. Then $\char R = p^n$ for some prime $p$, $n \in \mathbb{N}$.

Proof (for completeness): If $\char(R/\m) = 0$, then $\char R = 0$. If $\char(R/\m) = p$ for some prime $p$, then under the unique ring map $i : \Z \to R$, we have $i(p) \in \m$, so $i^{-1}(\m)$ is a prime ideal containing $p$, thus $i^{-1}(\m) = (p)$, and there is an induced map $\overline{i} : \Z_{(p)} \to R$. As $\Z_{(p)}$ is a DVR, every ideal of $\Z_{(p)}$ is of the form $p^n\Z_{(p)}$, so for some $n$, $\Z_{(p)}/p^n\Z_{(p)} \hookrightarrow R$, hence $\char R = \char(\Z_{(p)}/p^n\Z_{(p)}) = p^n$ (since characteristic may be computed in any subring).

Returning to the problem, suppose $R$ is a finite ring with $34$ units, and that $|R|$ is minimal among all examples. Then $R$ is Artinian, hence $R \cong \prod_{i=1}^n R_i$ is a finite product of Artinian local rings. Then $34 = |R^\times| = \prod_{i=1}^n |R_i^\times|$. Ignoring factors with $|R_i^\times| = 1$, we may assume $|R_i^\times| \ge 2$ for each $i$, which implies $n \le 2$. If $n = 2$, then WLOG $|R_1^\times| = 17$, but no finite ring has $17$ units (if a finite ring has an odd number of units, that number must be a product of terms of the form $2^k - 1$, see e.g. the linked paper below). Thus $n = 1$, so $R$ itself is Artinian local, with maximal ideal $\m$. Then $|R| = |\m| + |R^\times| = |\m| + 34 \implies \dfrac{34}{|\m|} = |R/\m| - 1 \in \Z$, so $|\m| = 1, 2, 17, 34$.

If $|\m| = 1$, then $\m = 0$, $R$ is a field, and $|R| = 1 + 34 = 35$, but no field has $35$ elements.

If $|\m| = 2$, then $|R/\m| = 17 + 1 = 18$, but no field has $18$ elements.

If $|\m| = 17$, then $|R| = 51$, so the additive group of $R$ is a group of order $51$. The only such group is the cyclic group of order $51$, so $(R, +) \cong \Z/51\Z$, hence $\char R = 51$, which contradicts the proposition.

If $|\m| = 34$, then $|R| = 68$, $|R/\m| = 2$, so $2 \mid \char R$. But the additive group of $R$ is an abelian group of order $68$, which has an element of order $17$, so $17 \mid \char R$ also, which again contradicts the proposition.

For much more in this regard, see this paper.