Let us use the convention that $\mathfrak{S}_0$ has a unique element $\sigma$ which is the identity (hence an involution) and has $\mathrm{tr}(\sigma)=0$. We also use $[n] := \{1,2,\ldots,n\}$.

The claimed identity is that for any $k\geq 0$,
$$ \sum_{n=0}^{\infty} \frac{z^n}{n!} \sum_{\sigma \in \mathrm{Inv}(\mathfrak{S}_n)}\mathrm{tr}(\sigma)^k = B_k(z)\cdot e^{z+\frac{1}{2}z^2}.$$

As mentioned, the case $k=0$ is well known. Hence we may assume $k \geq 1$ and that we want to prove,
$$ \sum_{n=1}^{\infty} \frac{z^n}{n!} \sum_{\sigma \in \mathrm{Inv}(\mathfrak{S}_n)}\mathrm{tr}(\sigma)^k = B_k(z)\cdot \sum_{n=0}^{\infty} \frac{z^n}{n!} \cdot \#\mathrm{Inv}(\mathfrak{S}_n).$$
Expanding coefficientwise the right-hand side, we see it is equal to
$$\sum_{n=0}^{\infty} \frac{z^n}{n!} \sum_{j=1}^{k} \frac{n!}{(n-j)!} \cdot S(k,j) \cdot \#\mathrm{Inv}(\mathfrak{S}_{n-j}).$$
Hence the equality we want is
$$ \sum_{\sigma \in \mathrm{Inv}(\mathfrak{S}_n)}\mathrm{tr}(\sigma)^k = \sum_{j=1}^{k} \frac{n!}{(n-j)!} \cdot S(k,j) \cdot \#\mathrm{Inv}(\mathfrak{S}_{n-j}).$$
We can prove this bijectively. The left-hand side counts pairs $(\sigma,(i_1,i_2,\ldots,i_k))$ with $\sigma \in \mathrm{Inv}(\mathfrak{S}_n)$ and $i_\ell$ a fixed-point of $\sigma$ for each $\ell=1,\ldots,k$ (allowing repeats). Suppose that $j := \#\{i_1,\ldots,i_k\}$ is the number of distinct entries among $(i_1,i_2,\ldots,i_k)$. Let $\widetilde{\sigma} \in \mathrm{Inv}(\mathfrak{S}_{n-j})$ be the permutation obtained from $\sigma$ by deleting the fixed-points in $\{i_1,i_2,\ldots,i_k\}$ and reindexing in an order-preserving way so the set being permuted is $[n-j]$. The map $(\sigma,(i_1,\ldots,i_k)) \to (\widetilde{\sigma},(i_1,\ldots,i_k))$ is clearly bijective. To finish the proof, I claim the number of possible sequences $(i_1,i_2,\ldots,i_k)$ which have $j := \#\{i_1,i_2,\ldots,i_k\}$ is $\frac{n}{(n-j)!} \cdot S(k,j) = \binom{n}{j} \cdot j! \cdot S(k,j)$. In fact, this is clear: there are $\binom{n}{j}$ choices for the subset $\{i_1,i_2,\ldots,i_k\} = \{\hat{i}_1 < \hat{i}_2 < \cdots <\hat{i}_j\}$, and $j! \cdot S(k,j)$ choices of a surjection $f\colon[k] \to [j]$, where our surjection is given by $f(a) =b$ if $i_a = \hat{i}_b$.