Do I use the Nernst equation when the concentrations of electrolyte in both half cells are equal?

I'm doing a chemistry report, and for my experiment I investigated the amount of voltage produced by different half cells in a voltaic cell.

However, I tested them under nonstandard conditions as the temperature at the time was $\pu{18^\circ C}$, and the concentration of the electrolytic solution in the half cells was $\pu{0.1 M}$.

I tried the Nernst equation, but since the concentration of both my electrolytes were $\pu{0.1 M}$, calculating the reaction quotient gave me an answer of $1$, and $\log 1 = 0$, therefore I'm left with the same cell potential I started with (cell potential at standard conditions) and I'm stuck there.

Here is an example for a magnesium-zinc voltaic cell at $\pu{291 K}$ ($\pu{18^\circ C}$):

$$E_\text{cell} = 1.593 - \frac{0.05778}{2} \log_{10} \frac{0.1}{0.1} = \pu{1.593 V}$$

Could you please explain the solution to me?

Yes, you still use the Nernst equation, but one of your implicit assumptions is not correct.

A point about the Nernst equation that often confuses people is that, at first glance, it doesn't predict any change in $E$ as temperature changes, as long as $Q$ remains equal to $1$, and $\ln Q = 0$.

$$E = E^\circ - \frac{RT}{\nu F}\ln Q \tag{1}$$

However, a slightly more careful analysis reveals that if you set $\ln Q = 0$, it only implies that $E = E^\circ$. The conclusion that $E$ doesn't change is based on an assumption that $E^\circ$ doesn't change, which is not true.

"Standard conditions" do not stipulate any temperature. So $E^\circ$ itself is a function of temperature. The standard cell potential $E^\circ$ at $\pu{298 K}$ is going to be different from the standard cell potential $E^\circ$ at $\pu{250 K}$, and hence $E$ will vary with temperature, even if $\ln Q = 0$.

So, how does one determine the temperature dependence of $E^\circ$? The best way is to measure it, but if that is not possible, then you have to somehow find an expression for it. The Nernst equation alone has no answer for this. Instead you would have to turn to thermodynamics. We know that

$$E^\circ = -\frac{\Delta_\mathrm rG^\circ}{\nu F} \tag{2}$$

Note that $\Delta_\mathrm rG^\circ$ is a function of temperature, and hence so is $E^\circ$. So, the problem becomes one of determining the variation of $\Delta_\mathrm rG^\circ$ with temperature. The most primitive way is probably to look up the data for $\Delta H_\mathrm{f}$ and $S_\mathrm{mol}$ of each of the compounds, calculate $\Delta_\mathrm{r} H^\circ$ and $\Delta_\mathrm{r} S^\circ$ for the reaction (using Hess's law), and then find

$$\Delta_\mathrm{r} G^\circ = \Delta_\mathrm{r} H^\circ - T\Delta_\mathrm{r} S^\circ \tag{3}$$

Of course, when you look up data, the chances are that the data you find are specified for $T = \pu{298 K}$. So, when you use these data to calculate $\Delta_\mathrm{r} H^\circ$ and $\Delta_\mathrm{r} S^\circ$, you are finding the values of these quantities at $T = \pu{298 K}$. If you plug these values into equation $(3)$, then you also assume that the values of $\Delta_\mathrm{r} H^\circ$ and $\Delta_\mathrm{r} S^\circ$ at your desired temperature are equal to their values at $\pu{298 K}$.

Depending on how much accuracy you need, this may or may not be tolerable, and there are certainly more thorough ways of calculating it. However, that discussion is best left for another question.