Can the change in internal energy be nonzero if temperature is constant?
TL;DR: Do not just memorise thermodynamics equations! And if you have an issue with the equations $\Delta U = 0$ or $\Delta H = 0$ for an isothermal process, read on...


The first problem
You said that an exothermic reaction corresponds to $\Delta U &lt; 0$. This is not true. It is defined by $\Delta H &lt; 0$.
However, we should still discuss this issue because $H$ is also a state function and in some cases $H$ is dependent only on temperature.

How to study thermodynamics
This kind of issue in thermodynamics frequently crops up here, and it is a very common mistake amongst students to indiscriminately use equations that they have learnt.
So, you have an equation that says $\Delta U = 0$ for an isothermal process, i.e. one at constant $T$. The important question here is not &quot;what is the equation&quot; or &quot;what is the answer&quot;! Instead, you should be asking yourself &quot;how do I derive this result&quot;, and the answer will naturally follow. This should really apply to everything you do - how can you expect to apply a formula that you do not actually understand?
Furthermore, if you are able to actually understand how an equation comes about, then you don't need to memorise it - you can derive it. For example, what's the point of memorising $w = -nRT\ln(V_2/V_1)$, along with the conditions, if you can simply obtain the equation from the very definition of work $đw = -\int p \,\mathrm{d}V$ by substituting in $p = nRT/V$ and integrating? What's more, the fact that you actually substituted in the ideal gas law should tell you something about the conditions that accompany the equations - for one, it's only applicable to ideal gases. What about when you reach the equation $\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V$ and happily memorise it, only to find out later that there are actually three more equations that look exactly like it? Are you going to memorise all four? No - you need to know where they come from.

Internal energy
In general, the internal energy $U$ of a substance is a function of at least two variables, for example $T$ and $p$. However, it can be shown using statistical mechanics that for an ideal gas,
$$U = n\left(\frac{3NRT}{2} + U_0\right)$$
Here, $N$ is the number of atoms in one molecule of the gas (for example, $N = 2$ for $\ce{H2}$, and $N = 3$ for $\ce{H2O}$). $U_0$ is the molar internal energy at absolute zero (this contains terms such as electronic energy), and crucially, this depends on the exact identity of the gas. This result is one example of the equipartition theorem in action (although the theorem itself is more powerful).
We do not need to concern ourselves with its derivation now. The point is that: this equation can only be applied to a fixed amount of a fixed ideal gas at a time. For example: if I have one mole of $\ce{H2}$ at $298\mathrm{~K}$, it is going to have a different internal energy from one mole of $\ce{Ar}$ at $298\mathrm{~K}$, even though the amount of substance and the temperature are the same.
What does this mean?
Firstly, it means that for anything that is not an ideal gas, you cannot assume that $U = U(T)$ and therefore you cannot assume that $\Delta T = 0$ implies $\Delta U = 0$. Let's say I compress $10\mathrm{~g}$ of water by reversibly increasing the pressure from $1\mathrm{~atm}$ to $10\mathrm{~atm}$, at a constant temperature of $323\mathrm{~K}$. Is the change in internal energy equal to zero? No, because water is not an ideal gas.
Secondly, it means that if the amount of ideal gas changes in any way, $U$ is going to change. Let's say I have a balloon filled with $1\mathrm{~mol}$ of argon gas, and I add a further $2\mathrm{~mol}$ argon gas into the balloon, all at a fixed temperature of $298\mathrm{~K}$. Is the change in internal energy equal to zero? No, in fact it will triple because you are changing $n$ from $1\mathrm{~mol}$ to $3\mathrm{~mol}$!
Lastly, it means that if the chemical composition is changing in any way, you are automatically not allowed to say that $\Delta T = 0$ implies $\Delta U = 0$. And this is the case even if all reactants and products are ideal gases. Why not? Let's look at this reaction, and let's assume that everything there behaves as an ideal gas, and let's say that the reaction vessel is kept at a constant temperature of $300\mathrm{~K}$.
$$\ce{H2 (g) + Cl2 (g) -&gt; 2HCl (g)}$$
Let's write an expression for the internal energy of the reactants. On the left-hand side, we have:
$$U_\mathrm{reactants} = \left[\frac{3}{2}RT + U_0(\ce{H2})\right] + \left[\frac{3}{2}RT + U_0(\ce{Cl2})\right]$$
On the right-hand side, we have:
$$U_\mathrm{products} = 3RT + 2U_0(\ce{HCl})$$
So, is $\Delta U$ equal to zero? No; even though the multiples of $RT$ cancel out with each other, the values of $U_0$ for the reacting species are different, which makes
$$\Delta U = 2U_0(\ce{HCl}) - U_0(\ce{H2}) - U_0(\ce{Cl2}) \neq 0.$$

Enthalpy
Let's go back to our (closed) system where there is only one ideal gas and it's not undergoing any chemical reaction. Within this system, we can safely say that if $\Delta T = 0$, then $\Delta U = 0$. So:
$$\begin{align}
\Delta H &amp;= \Delta (U + pV) \\
&amp;= \Delta U + \Delta (pV) \\
&amp;= \Delta (pV) \qquad \\
&amp;= \Delta (nRT) \\
&amp;= nR \Delta T \\
&amp;= 0
\end{align}$$
We haven't made any assumptions here apart from the ideality of the gas (the fact that $n$ is constant came from the fact that it is a closed system). So, for an ideal gas, $\Delta T = 0$ implies $\Delta H = 0$.
However, we did use the fact that $\Delta U = 0$. Therefore, what I said about the cases above where $\Delta U \neq 0$ also applies here.
Regarding the terms endothermic and exothermic, they are used to describe chemical reactions. And for chemical reactions, there is absolutely no way you can say that constant temperature implies $\Delta U = 0$ or $\Delta H = 0$. That's why we can actually use the terms endothermic and exothermic. If every single chemical reaction had $\Delta H = 0$, life would be rather plain!