How can I derive the mean and variance of a $k$-truncated Poisson? Here, $k$ is the cutoff value such that only values strictly larger than $k$ are allowed, i.e. the probability mass function is
$$p_j = q_k^{-1} \lambda^j e^{-\lambda}/j!, \qquad j=k+1, k+2, \dots$$
where $k \geq 0$ is an integer, $\lambda \gt 0$ is a parameter, and $q_k = 1 - \sum_{i=0}^{k} \lambda^i e^{-\lambda}/i!$.
Derivation of mean
Here is the ugly, brute force method. Recall that the mean of $X \sim \operatorname{Poisson}(\lambda)$ can be found as follows:
$$\mathbb{E}(X) = \sum_{x=0}^{\infty} x p_x = \sum_{x=0}^{\infty} x \left( \frac{\lambda^x e^{-\lambda}}{x!} \right) $$
In particular observe that the $x=0$ term makes no contribution to this sum so can be dropped (we start the sum at $x=1$ instead), and that for $x>0$ we can cancel $x$ and $x!$ to leave $(x-1)!$ in the denominator:
$$\mathbb{E}(X) = \sum_{x=1}^{\infty} \frac{\lambda^x e^{-\lambda}}{(x-1)!} $$
We observe this summand closely resembles a Poisson PMF again, but would prefer to see $x-1$ in the exponent of $\lambda$. We can achieve this by factoring a $\lambda$ outside the sum. For convenience, relabel $y=x-1$:
$$\mathbb{E}(X) = \lambda \sum_{x=1}^{\infty} \frac{\lambda^{x-1} e^{-\lambda}}{(x-1)!} = \lambda \sum_{y=0}^{\infty} \frac{\lambda^y e^{-\lambda}}{y!} = \lambda \sum_{y=0}^{\infty} p_y = \lambda (1) = \lambda$$
What difference does the $k$-truncated distribution make? Our first sum need only start at $x=k+1$, since $p_x = 0$ for $x \leq k$, and the PMF has an extra factor of $q_k^{-1}$ which we may as well factor outside. We again factor a $\lambda$ and set $y=x-1$ to obtain a Poisson PMF:
$$\mathbb{E}(X) = \sum_{x=k+1}^{\infty} x p_x = q_k^{-1} \sum_{x=k+1}^{\infty} x \left( \frac{\lambda^x e^{-\lambda}}{x!} \right) = \lambda q_k^{-1} \sum_{x=k+1}^{\infty} \frac{\lambda^{x-1} e^{-\lambda}}{(x-1)!} = \lambda q_k^{-1} \sum_{y=k}^{\infty} \frac{\lambda^{y} e^{-\lambda}}{y!}$$
This sum does not come to unity, but $\sum_{y=k}^{\infty} \lambda^{y} e^{-\lambda}/y! = 1 - \sum_{y=0}^{k-1} \lambda^{y} e^{-\lambda}/y! = q_{k-1}$ (where, in the special case that $k=0$, we understand $q_{-1} = 1$).
Therefore we obtain:
$$\mathbb{E}(X) = \frac {\lambda q_{k-1}}{q_k}$$
Derivation of variance
We can do something similar to work out the variance, for which it is easier to use factorial moments to help with the factorial-cancelling trick. The second factorial moment of a Poisson distribution is:
$$\mathbb{E}\left((X)_2\right) = \mathbb{E}\left(X(X-1)\right) = \sum_{x=0}^{\infty} x(x-1) p_x = \sum_{x=0}^{\infty} x(x-1) \left( \frac{\lambda^x e^{-\lambda}}{x!} \right) $$
Similar to how we calculate the mean, we note that the first two terms make no contribution to the sum and can be neglected, so that our starting index is $x=2$, and for $x \geq 2$ we may cancel our falling factorial $x(x-1)$ with the $x!$ in the denominator. To obtain a matching exponent of the $\lambda$ we factor out $\lambda^2$, and analogous to before we can set $y=x-2$:
$$\mathbb{E}\left((X)_2\right) = \lambda^2 \sum_{x=2}^{\infty} \frac{\lambda^{x-2} e^{-\lambda}}{(x-2)!} = \lambda^2 \sum_{y=0}^{\infty} \frac{\lambda^{y} e^{-\lambda}}{y!} =\lambda^2 (1) = \lambda^2 \tag{1}$$
Since $\mathbb{E}\left((X)_2\right) = \mathbb{E}\left(X(X-1)\right) = \mathbb{E}(X^2) - \mathbb{E}(X)$ and $\operatorname{Var}(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2$, we have the general rule $\operatorname{Var}(X) = \mathbb{E}\left((X)_2\right) + \mathbb{E}(X) - \mathbb{E}(X)^2$. Specifically for the Poisson distribution, we therefore obtain $\operatorname{Var}(X) = \lambda^2 + \lambda - \lambda^2 = \lambda$.
Again, we attempt to repeat this analysis for the $k$-truncated distribution.
$$\mathbb{E}\left((X)_2\right) = \sum_{x=k+1}^{\infty} x(x-1) p_x = q_k^{-1} \sum_{x=k+1}^{\infty} x(x-1) \left( \frac{\lambda^x e^{-\lambda}}{x!} \right) \tag{2} $$
Assuming $k \geq 1$, we can cancel the factorial as before to produce:
$$\mathbb{E}\left((X)_2\right) = \lambda^2 q_k^{-1} \sum_{x=k+1}^{\infty} \frac{\lambda^{x-2} e^{-\lambda}}{(x-2)!} = \lambda^2 q_k^{-1} \sum_{y=k-1}^{\infty} \frac{\lambda^{y} e^{-\lambda}}{y!} = \frac{ \lambda^2 q_{k-2}}{q_k}$$
We need to reason separately about the case $k=0$ in $(2)$, since the first term of the sum, where $x=1$, will then be $(1)(0)\frac{\lambda e^{-\lambda}}{1!}$ and we can't cancel the factorial to $(-1)!$. However, this term is zero so we can commence the summation at $x=2$ instead. This is exactly the sum we performed for the untruncated Poisson distribution in $(1)$, so we obtain $\mathbb{E}(X) = q_k^{-1} \lambda^2$. If we take $q_{-2}=1$ then we can use $\lambda^2 q_{k-2}/q_k$ for $k=0$ as well as $k \geq 1$; there is no need to treat $k=0$ as a special case in our final formula.
Applying $\operatorname{Var}(X) = \mathbb{E}\left((X)_2\right) + \mathbb{E}(X) - \mathbb{E}(X)^2$ again, we obtain:
$$\operatorname{Var}(X) = \frac{\lambda^2\left( q_{k-2}q_k - q_{k-1}^2 \right)}{q_k^2} + \frac{\lambda q_{k-1}}{q_k}$$
Comments
As we have seen, it is helpful to define $q_k = 1$ for $k < 0$, which was not covered explicitly in the original question. It is reassuring to discover that setting $k=-1$ in our formulae we recover the mean and variance of the untruncated Poisson. The factorial moments method extends well for higher moments; for the untruncated Poisson we have $\mathbb{E}\left((X)_r\right) = \lambda^r$. More sophisticated methods exist but I wanted to present present an approach that only assumes the reader to be familiar with the basic machinery of mean and variance calculations.
A quick numerical simulation in R
k <- 10
lambda <- 3
n <- 30 #highest index to go up to
qk <- function(k) {1 - ppois(k, lambda=lambda)}
pj <- sapply(1:n, function(x) {
ifelse(x > k, qk(k)^(-1) * dpois(x, lambda=lambda), 0)})
> sum(pj) #verify total probability comes to 1
[1] 1
> sum(pj * 1:n) #simulated mean
[1] 11.31388
> lambda * qk(k-1) / qk(k) #mean from formula
[1] 11.31388
> sum(pj * (1:n)^2) - sum(pj * 1:n)^2 #simulated variance
[1] 0.3904626
> lambda^2 * (qk(k-2)*qk(k) - qk(k-1)^2) / qk(k)^2 + lambda * qk(k-1) / qk(k) #variance from formula
[1] 0.3904626