Suppose the lens has an aperture radius of r and focal length f. The image of an object at infinity will then appear at a distance f from the lens. The angular separation between two points of the object is then the same as the angle between the image of the two points in the focal point seen from a distance of $f$. The area of the image of the object is thus given by $\pi\alpha^2 f^2$ where $\alpha$ is the angle between the center of the object (assumed to be spherical) and the edge, if we assume that this angle is small.
If the object radiates as a black body, has a radius of R, a temperature of $T$ and is a distance d away, then the flux of radiation reaching the lens is:
$$F = \sigma T^4 \left(\frac{R}{d}\right)^2 = \sigma T^4 \alpha^2$$
where $\sigma$ is the Stefan–Boltzmann constant. The total power of the radiation entering the lens $P$ is the area of the lens opening times the flux:
$$P = \pi r^2 F = \pi \sigma T^4\alpha^2r^2$$
This power ends up heating the area of the image in the focal plane. The flux of radiation there is:
$$F_{\text{im}} = \frac{P}{\pi\alpha^2f^2} = \sigma T^4\frac{r^2}{f^2}$$
Suppose then that you put a black body in the image plane, then the temperature there would be $T_{\text{im}}$ where $\sigma T_{\text{im}}^4 = F_{\text{im}}$, therefore:
$$T_{\text{im}} = \sqrt{\frac{r}{f}}T$$
The ratio of the focal length f and the lens diameter is called the F-number and this is always larger than 1. So, the factor multiplying $T$ in the above equation will always be smaller than 1, therefore you can never reach a higher temperature than the temperature of the object in this way.
